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Conic Sections
Identify the equation of a parabola in standard form with given focus and directrix.
Parabolas
A parabola is generated when a plane intersects a cone parallel to the generating line. In this case, the plane intersects only one of the nappes. A parabola can also be defined in terms of distances.
A graph of a typical parabola appears in . Using this diagram in conjunction with the distance formula, we can derive an equation for a parabola. Recall the distance formula: Given point P with coordinates \(({x}_{1},{y}_{1})\) and point Q with coordinates \(({x}_{2},{\ \text{y}}_{2}),\) the distance between them is given by the formula
\[d(P,Q)=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}.\]Then from the definition of a parabola and , we get
\[\begin{array}{lll}d(F,P) & = & d(P,Q) \\ \sqrt{{(0-x)}^{2}+{(p-y)}^{2}} & = & \sqrt{{(x-x)}^{2}+{(\text{-}p-y)}^{2}}.\end{array}\]Squaring both sides and simplifying yields
\[\begin{array}{lll}{x}^{2}+{(p-y)}^{2} & = & {0}^{2}+{(\text{-}p-y)}^{2} \\ {x}^{2}+{p}^{2}-2py+{y}^{2} & = & {p}^{2}+2py+{y}^{2} \\ {x}^{2}-2py & = & 2py \\ {x}^{2} & = & 4py.\end{array}\]Now suppose we want to relocate the vertex. We use the variables \((h,k)\) to denote the coordinates of the vertex. Then if the focus is directly above the vertex, it has coordinates \((h,k+p)\) and the directrix has the equation \(y=k-p.\) Going through the same derivation yields the formula \({(x-h)}^{2}=4p(y-k).\) Solving this equation for y leads to the following theorem.
We can also study the cases when the parabola opens down or to the left or the right. The equation for each of these cases can also be written in standard form as shown in the following graphs.
In addition, the equation of a parabola can be written in the general form, though in this form the values of h, k, and p are not immediately recognizable. The general form of a parabola is written as
\[a{x}^{2}+bx+cy+d=0\ \text{or}\ a{y}^{2}+bx+cy+d=0.\]Condensed — the full section is in OpenStax Calculus Volume 3.
Ellipses
An ellipse can also be defined in terms of distances. In the case of an ellipse, there are two foci (plural of focus), and two directrices (plural of directrix). We look at the directrices in more detail later in this section.
A graph of a typical ellipse is shown in . In this figure the foci are labeled as \(F\) and \({F}^{'}.\) Both are the same fixed distance from the origin, and this distance is represented by the variable c. Therefore the coordinates of \(F\) are \((c,0)\) and the coordinates of \({F}^{'}\) are \((\text{-}c,0).\) The points \(P\) and \({P}^{'}\) are located at the ends of the major axis of the ellipse, and have coordinates \((a,0)\) and \((\text{-}a,0),\) respectively. The major axis is always the longest distance across the ellipse, and can be horizontal or vertical. Thus, the length of the major axis in this ellipse is 2a. Furthermore, \(P\) and \({P}^{'}\) are called the vertices of the ellipse. The points \(Q\) and \({Q}^{'}\) are located at the ends of the minor axis of the ellipse, and have coordinates \((0,b)\) and \((0,\text{-}b),\) respectively. The minor axis is the shortest distance across the ellipse. The minor axis is perpendicular to the major axis.
According to the definition of the ellipse, we can choose any point on the ellipse and the sum of the distances from this point to the two foci is constant. Suppose we choose the point P. Since the coordinates of point P are \((a,0),\) the sum of the distances is
\[d(P,F)+d(P,{F}^{'})=(a-c)+(a+c)=2a.\]Therefore the sum of the distances from an arbitrary point A with coordinates \((x,y)\) is also equal to 2a. Using the distance formula, we get
\[\begin{array}{lll}d(A,F)+d(A,{F}^{'}) & = & 2a \\ \sqrt{{(x-c)}^{2}+{y}^{2}}+\sqrt{{(x+c)}^{2}+{y}^{2}} & = & 2a.\end{array}\]Subtract the second radical from both sides and square both sides:
\[\begin{array}{lll}\sqrt{{(x-c)}^{2}+{y}^{2}} & = & 2a-\sqrt{{(x+c)}^{2}+{y}^{2}} \\ {(x-c)}^{2}+{y}^{2} & = & 4{a}^{2}-4a\sqrt{{(x+c)}^{2}+{y}^{2}}+{(x+c)}^{2}+{y}^{2} \\ {x}^{2}-2cx+{c}^{2}+{y}^{2} & = & 4{a}^{2}-4a\sqrt{{(x+c)}^{2}+{y}^{2}}+{x}^{2}+2cx+{c}^{2}+{y}^{2} \\ \text{-}2cx & = & 4{a}^{2}-4a\sqrt{{(x+c)}^{2}+{y}^{2}}+2cx.\end{array}\]Now isolate the radical on the right-hand side and square again:
Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:
\[\begin{array}{lll} \\ {x}^{2}-\frac{{c}^{2}{x}^{2}}{{a}^{2}}+{y}^{2} & = & {a}^{2}-{c}^{2} \\ \frac{({a}^{2}-{c}^{2}){x}^{2}}{{a}^{2}}+{y}^{2} & = & {a}^{2}-{c}^{2}.\end{array}\]Condensed — the full section is in OpenStax Calculus Volume 3.
Hyperbolas
A hyperbola can also be defined in terms of distances. In the case of a hyperbola, there are two foci and two directrices. Hyperbolas also have two asymptotes.
A graph of a typical hyperbola appears as follows.
The derivation of the equation of a hyperbola in standard form is virtually identical to that of an ellipse. One slight hitch lies in the definition: The difference between two numbers is always positive. Let P be a point on the hyperbola with coordinates \((x,y).\) Then the definition of the hyperbola gives \(|d(P,{F}_{1})-d(P,{F}_{2})|=\text{constant}.\) To simplify the derivation, assume that P is on the right branch of the hyperbola, so the absolute value bars drop. If it is on the left branch, then the subtraction is reversed. The vertex of the right branch has coordinates \((a,0),\) so
\[d(P,{F}_{1})-d(P,{F}_{2})=(c+a)-(c-a)=2a.\]This equation is therefore true for any point on the hyperbola. Returning to the coordinates \((x,y)\) for P:
\[\begin{array}{lll}d(P,{F}_{1})-d(P,{F}_{2}) & = & 2a \\ \sqrt{{(x+c)}^{2}+{y}^{2}}-\sqrt{{(x-c)}^{2}+{y}^{2}} & = & 2a.\end{array}\]Add the second radical from both sides and square both sides:
\[\begin{array}{lll}\sqrt{{(x-c)}^{2}+{y}^{2}} & = & 2a+\sqrt{{(x+c)}^{2}+{y}^{2}} \\ {(x-c)}^{2}+{y}^{2} & = & 4{a}^{2}+4a\sqrt{{(x+c)}^{2}+{y}^{2}}+{(x+c)}^{2}+{y}^{2} \\ {x}^{2}-2cx+{c}^{2}+{y}^{2} & = & 4{a}^{2}+4a\sqrt{{(x+c)}^{2}+{y}^{2}}+{x}^{2}+2cx+{c}^{2}+{y}^{2} \\ \text{-}2cx & = & 4{a}^{2}+4a\sqrt{{(x+c)}^{2}+{y}^{2}}+2cx.\end{array}\]Now isolate the radical on the right-hand side and square again:
\[\begin{array}{lll}-2cx & = & 4{a}^{2}+4a\sqrt{{(x+c)}^{2}+{y}^{2}}+2cx \\ 4a\sqrt{{(x+c)}^{2}+{y}^{2}} & = & -4{a}^{2}-4cx \\ \sqrt{{(x+c)}^{2}+{y}^{2}} & = & \text{-}a-\frac{cx}{a} \\ {(x+c)}^{2}+{y}^{2} & = & {a}^{2}+2cx+\frac{{c}^{2}{x}^{2}}{{a}^{2}} \\ {x}^{2}+2cx+{c}^{2}+{y}^{2} & = & {a}^{2}+2cx+\frac{{c}^{2}{x}^{2}}{{a}^{2}} \\ {x}^{2}+{c}^{2}+{y}^{2} & = & {a}^{2}+\frac{{c}^{2}{x}^{2}}{{a}^{2}}.\end{array}\]Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:
\[\begin{array}{lll} \\ {x}^{2}-\frac{{c}^{2}{x}^{2}}{{a}^{2}}+{y}^{2} & = & {a}^{2}-{c}^{2} \\ \frac{({a}^{2}-{c}^{2}){x}^{2}}{{a}^{2}}+{y}^{2} & = & {a}^{2}-{c}^{2}.\end{array}\]\[\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{a}^{2}-{c}^{2}}=1.\]\[\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1.\]Condensed — the full section is in OpenStax Calculus Volume 3.
Eccentricity and Directrix
An alternative way to describe a conic section involves the directrices, the foci, and a new property called eccentricity. We will see that the value of the eccentricity of a conic section can uniquely define that conic.
The three conic sections with their directrices appear in the following figure.
Recall from the definition of a parabola that the distance from any point on the parabola to the focus is equal to the distance from that same point to the directrix. Therefore, by definition, the eccentricity of a parabola must be 1. The equations of the directrices of a horizontal ellipse are \(x=\text{\pm }\frac{{a}^{2}}{c}.\) The right vertex of the ellipse is located at \((a,0)\) and the right focus is \((c,0).\) Therefore the distance from the vertex to the focus is \(a-c\) and the distance from the vertex to the right directrix is \(\frac{{a}^{2}}{c}-a.\) This gives the eccentricity as
\[e=\frac{a-c}{\frac{{a}^{2}}{c}-a}=\frac{c(a-c)}{{a}^{2}-ac}=\frac{c(a-c)}{a(a-c)}=\frac{c}{a}.\]Since \(c
Example
Try it.
Determine the eccentricity of the ellipse described by the equation
\[\frac{{(x-3)}^{2}}{16}+\frac{{(y+2)}^{2}}{25}=1.\]Solution
From the equation we see that \(a=5\) and \(b=4.\) The value of c can be calculated using the equation \({a}^{2}={b}^{2}+{c}^{2}\) for an ellipse. Substituting the values of a and b and solving for c gives \(c=3.\) Therefore the eccentricity of the ellipse is \(e=\frac{c}{a}=\frac{3}{5}=0.6.\)
Polar Equations of Conic Sections
Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter. The focal parameter of a conic section p is defined as the distance from a focus to the nearest directrix. The following table gives the focal parameters for the different types of conics, where a is the length of the semi-major axis (i.e., half the length of the major axis), c is the distance from the origin to the focus, and e is the eccentricity. In the case of a parabola, a represents the distance from the vertex to the focus.
| Conic | e | p |
| Ellipse | \(0| \(\frac{{a}^{2}-{c}^{2}}{c}=\frac{a(1-{e}^{2})}{e}\) | |
| Parabola | \(e=1\) | \(2a\) |
| Hyperbola | \(e>1\) | \(\frac{{c}^{2}-{a}^{2}}{c}=\frac{a({e}^{2}-1)}{e}\) |
Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conic section in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then using the definition of the various conic sections in terms of distances, it is possible to prove the following theorem.
In the equation on the left, the major axis of the conic section is horizontal, and in the equation on the right, the major axis is vertical. To work with a conic section written in polar form, first make the constant term in the denominator equal to 1. This can be done by dividing both the numerator and the denominator of the fraction by the constant that appears in front of the plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. This value identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conic is vertical. If both appear then the axes are rotated. The center of the conic is not necessarily at the origin. The center is at the origin only if the conic is a circle (i.e., \(e=0).\)
Condensed — the full section is in OpenStax Calculus Volume 3.
General Equations of Degree Two
A general equation of degree two can be written in the form
\[A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0.\]The graph of an equation of this form is a conic section. If \(B\ne 0\) then the coordinate axes are rotated. To identify the conic section, we use the discriminant of the conic section \(4AC-{B}^{2}.\) One of the following cases must be true:
- \(4AC-{B}^{2}>0.\) If so, the graph is an ellipse.
- \(4AC-{B}^{2}=0.\) If so, the graph is a parabola.
- \(4AC-{B}^{2}<0.\) If so, the graph is a hyperbola.
The simplest example of a second-degree equation involving a cross term is \(xy=1.\) This equation can be solved for y to obtain \(y=\frac{1}{x}.\) The graph of this function is called a rectangular hyperbola as shown.
The asymptotes of this hyperbola are the x and y coordinate axes. To determine the angle \(\theta\) of rotation of the conic section, we use the formula \(\text{cot}\ 2\theta =\frac{A-C}{B}.\) In this case \(A=C=0\) and \(B=1,\) so \(\text{cot}\ 2\theta =(0-0)\text{/}1=0\) and \(\theta =45\text{^{\circ}}.\) The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled \({A}^{'},{B}^{'},{C}^{'},{D}^{'},{E}^{'},\ \text{and}\ {F}^{'},\) and are given by the formulas
\[\begin{array}{lll}{A}^{'} & = & A\ {\text{cos}}^{2}\theta +B\ \text{cos}\ \theta \ \text{sin}\ \theta +C\ {\text{sin}}^{2}\theta \\ {B}^{'} & = & 0 \\ {C}^{'} & = & A\ {\text{sin}}^{2}\theta -B\ \text{sin}\ \theta \ \text{cos}\ \theta +C\ {\text{cos}}^{2}\theta \\ {D}^{'} & = & D\ \text{cos}\ \theta +E\ \text{sin}\ \theta \\ {E}^{'} & = & \text{-}D\ \text{sin}\ \theta +E\ \text{cos}\ \theta \\ {F}^{'} & = & F.\end{array}\]The procedure for graphing a rotated conic is the following:
- Identify the conic section using the discriminant \(4AC-{B}^{2}.\)
- Determine \(\theta\) using the formula \(\text{cot}\ 2\theta =\frac{A-C}{B}.\)
- Calculate \({A}^{'},{B}^{'},{C}^{'},{D}^{'},{E}^{'},\ \text{and}\ {F}^{'}.\)
- Rewrite the original equation using \({A}^{'},{B}^{'},{C}^{'},{D}^{'},{E}^{'},\ \text{and}\ {F}^{'}.\)
- Draw a graph using the rotated equation.
Condensed — the full section is in OpenStax Calculus Volume 3.
Key Concepts
- The equation of a vertical parabola in standard form with given focus and directrix is \(y=\frac{1}{4p}{(x-h)}^{2}+k\) where p is the distance from the vertex to the focus and \((h,k)\) are the coordinates of the vertex.
- The equation of a horizontal ellipse in standard form is \(\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1\) where the center has coordinates \((h,k),\) the major axis has length 2a, the minor axis has length 2b, and the coordinates of the foci are \((h\pm c,k),\) where \({c}^{2}={a}^{2}-{b}^{2}.\)
- The equation of a horizontal hyperbola in standard form is \(\frac{{(x-h)}^{2}}{{a}^{2}}-\frac{{(y-k)}^{2}}{{b}^{2}}=1\) where the center has coordinates \((h,k),\) the vertices are located at \((h\pm a,k),\) and the coordinates of the foci are \((h\pm c,k),\) where \({c}^{2}={a}^{2}+{b}^{2}.\)
- The eccentricity of an ellipse is less than 1, the eccentricity of a parabola is equal to 1, and the eccentricity of a hyperbola is greater than 1. The eccentricity of a circle is 0.
- The polar equation of a conic section with eccentricity e is \(r=\frac{ep}{1\pm e\ \text{cos}\ \theta }\) or \(r=\frac{ep}{1\pm e\ \text{sin}\ \theta },\) where p represents the focal parameter.
- To identify a conic generated by the equation \(A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0,\) first calculate the discriminant \(D=4AC-{B}^{2}.\) If \(D>0\) then the conic is an ellipse, if \(D=0\) then the conic is a parabola, and if \(D<0\) then the conic is a hyperbola.
Conic Sections
For the following exercises, determine the equation of the parabola using the information given.
For the following exercises, determine the equation of the ellipse using the information given.
For the following exercises, determine the equation of the hyperbola using the information given.
For the following exercises, consider the following polar equations of conics. Determine the eccentricity and identify the conic.
For the following exercises, find a polar equation of the conic with focus at the origin and eccentricity and directrix as given.
For the following exercises, sketch the graph of each conic.
For the following equations, determine which of the conic sections is described.
Condensed — the full section is in OpenStax Calculus Volume 3.
Parabolas
A parabola is generated when a plane intersects a cone parallel to the generating line. In this case, the plane intersects only one of the nappes. A parabola can also be defined in terms of distances.
A graph of a typical parabola appears in . Using this diagram in conjunction with the distance formula, we can derive an equation for a parabola. Recall the distance formula: Given point P with coordinates \(({x}_{1},{y}_{1})\) and point Q with coordinates \(({x}_{2},{\ \text{y}}_{2}),\) the distance between them is given by the formula
\[d(P,Q)=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}.\]Then from the definition of a parabola and , we get
\[\begin{array}{lll}d(F,P) & = & d(P,Q) \\ \sqrt{{(0-x)}^{2}+{(p-y)}^{2}} & = & \sqrt{{(x-x)}^{2}+{(\text{-}p-y)}^{2}}.\end{array}\]Squaring both sides and simplifying yields
\[\begin{array}{lll}{x}^{2}+{(p-y)}^{2} & = & {0}^{2}+{(\text{-}p-y)}^{2} \\ {x}^{2}+{p}^{2}-2py+{y}^{2} & = & {p}^{2}+2py+{y}^{2} \\ {x}^{2}-2py & = & 2py \\ {x}^{2} & = & 4py.\end{array}\]Now suppose we want to relocate the vertex. We use the variables \((h,k)\) to denote the coordinates of the vertex. Then if the focus is directly above the vertex, it has coordinates \((h,k+p)\) and the directrix has the equation \(y=k-p.\) Going through the same derivation yields the formula \({(x-h)}^{2}=4p(y-k).\) Solving this equation for y leads to the following theorem.
We can also study the cases when the parabola opens down or to the left or the right. The equation for each of these cases can also be written in standard form as shown in the following graphs.
In addition, the equation of a parabola can be written in the general form, though in this form the values of h, k, and p are not immediately recognizable. The general form of a parabola is written as
\[a{x}^{2}+bx+cy+d=0\ \text{or}\ a{y}^{2}+bx+cy+d=0.\]Condensed — the full section is in OpenStax Calculus Volume 2.
Ellipses
An ellipse can also be defined in terms of distances. In the case of an ellipse, there are two foci (plural of focus), and two directrices (plural of directrix). We look at the directrices in more detail later in this section.
A graph of a typical ellipse is shown in . In this figure the foci are labeled as \(F\) and \({F}^{'}.\) Both are the same fixed distance from the origin, and this distance is represented by the variable c. Therefore the coordinates of \(F\) are \((c,0)\) and the coordinates of \({F}^{'}\) are \((\text{-}c,0).\) The points \(P\) and \({P}^{'}\) are located at the ends of the major axis of the ellipse, and have coordinates \((a,0)\) and \((\text{-}a,0),\) respectively. The major axis is always the longest distance across the ellipse, and can be horizontal or vertical. Thus, the length of the major axis in this ellipse is 2a. Furthermore, \(P\) and \({P}^{'}\) are called the vertices of the ellipse. The points \(Q\) and \({Q}^{'}\) are located at the ends of the minor axis of the ellipse, and have coordinates \((0,b)\) and \((0,\text{-}b),\) respectively. The minor axis is the shortest distance across the ellipse. The minor axis is perpendicular to the major axis.
According to the definition of the ellipse, we can choose any point on the ellipse and the sum of the distances from this point to the two foci is constant. Suppose we choose the point P. Since the coordinates of point P are \((a,0),\) the sum of the distances is
\[d(P,F)+d(P,{F}^{'})=(a-c)+(a+c)=2a.\]Therefore the sum of the distances from an arbitrary point A with coordinates \((x,y)\) is also equal to 2a. Using the distance formula, we get
\[\begin{array}{lll}d(A,F)+d(A,{F}^{'}) & = & 2a \\ \sqrt{{(x-c)}^{2}+{y}^{2}}+\sqrt{{(x+c)}^{2}+{y}^{2}} & = & 2a.\end{array}\]Subtract the second radical from both sides and square both sides:
\[\begin{array}{lll}\sqrt{{(x-c)}^{2}+{y}^{2}} & = & 2a-\sqrt{{(x+c)}^{2}+{y}^{2}} \\ {(x-c)}^{2}+{y}^{2} & = & 4{a}^{2}-4a\sqrt{{(x+c)}^{2}+{y}^{2}}+{(x+c)}^{2}+{y}^{2} \\ {x}^{2}-2cx+{c}^{2}+{y}^{2} & = & 4{a}^{2}-4a\sqrt{{(x+c)}^{2}+{y}^{2}}+{x}^{2}+2cx+{c}^{2}+{y}^{2} \\ \text{-}2cx & = & 4{a}^{2}-4a\sqrt{{(x+c)}^{2}+{y}^{2}}+2cx.\end{array}\]Now isolate the radical on the right-hand side and square again:
Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:
\[\begin{array}{lll} \\ {x}^{2}-\frac{{c}^{2}{x}^{2}}{{a}^{2}}+{y}^{2} & = & {a}^{2}-{c}^{2} \\ \frac{({a}^{2}-{c}^{2}){x}^{2}}{{a}^{2}}+{y}^{2} & = & {a}^{2}-{c}^{2}.\end{array}\]Condensed — the full section is in OpenStax Calculus Volume 2.
Hyperbolas
A hyperbola can also be defined in terms of distances. In the case of a hyperbola, there are two foci and two directrices. Hyperbolas also have two asymptotes.
A graph of a typical hyperbola appears as follows.
The derivation of the equation of a hyperbola in standard form is virtually identical to that of an ellipse. One slight hitch lies in the definition: The difference between two numbers is always positive. Let P be a point on the hyperbola with coordinates \((x,y).\) Then the definition of the hyperbola gives \(|d(P,{F}_{1})-d(P,{F}_{2})|=\text{constant}.\) To simplify the derivation, assume that P is on the right branch of the hyperbola, so the absolute value bars drop. If it is on the left branch, then the subtraction is reversed. The vertex of the right branch has coordinates \((a,0),\) so
\[d(P,{F}_{1})-d(P,{F}_{2})=(c+a)-(c-a)=2a.\]This equation is therefore true for any point on the hyperbola. Returning to the coordinates \((x,y)\) for P:
\[\begin{array}{lll}d(P,{F}_{1})-d(P,{F}_{2}) & = & 2a \\ \sqrt{{(x+c)}^{2}+{y}^{2}}-\sqrt{{(x-c)}^{2}+{y}^{2}} & = & 2a.\end{array}\]Add the second radical from both sides and square both sides:
\[\begin{array}{lll}\sqrt{{(x-c)}^{2}+{y}^{2}} & = & 2a+\sqrt{{(x+c)}^{2}+{y}^{2}} \\ {(x-c)}^{2}+{y}^{2} & = & 4{a}^{2}+4a\sqrt{{(x+c)}^{2}+{y}^{2}}+{(x+c)}^{2}+{y}^{2} \\ {x}^{2}-2cx+{c}^{2}+{y}^{2} & = & 4{a}^{2}+4a\sqrt{{(x+c)}^{2}+{y}^{2}}+{x}^{2}+2cx+{c}^{2}+{y}^{2} \\ \text{-}2cx & = & 4{a}^{2}+4a\sqrt{{(x+c)}^{2}+{y}^{2}}+2cx.\end{array}\]Now isolate the radical on the right-hand side and square again:
\[\begin{array}{lll}-2cx & = & 4{a}^{2}+4a\sqrt{{(x+c)}^{2}+{y}^{2}}+2cx \\ 4a\sqrt{{(x+c)}^{2}+{y}^{2}} & = & -4{a}^{2}-4cx \\ \sqrt{{(x+c)}^{2}+{y}^{2}} & = & \text{-}a-\frac{cx}{a} \\ {(x+c)}^{2}+{y}^{2} & = & {a}^{2}+2cx+\frac{{c}^{2}{x}^{2}}{{a}^{2}} \\ {x}^{2}+2cx+{c}^{2}+{y}^{2} & = & {a}^{2}+2cx+\frac{{c}^{2}{x}^{2}}{{a}^{2}} \\ {x}^{2}+{c}^{2}+{y}^{2} & = & {a}^{2}+\frac{{c}^{2}{x}^{2}}{{a}^{2}}.\end{array}\]Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:
\[\begin{array}{lll} \\ {x}^{2}-\frac{{c}^{2}{x}^{2}}{{a}^{2}}+{y}^{2} & = & {a}^{2}-{c}^{2} \\ \frac{({a}^{2}-{c}^{2}){x}^{2}}{{a}^{2}}+{y}^{2} & = & {a}^{2}-{c}^{2}.\end{array}\]\[\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{a}^{2}-{c}^{2}}=1.\]\[\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1.\]Condensed — the full section is in OpenStax Calculus Volume 2.
Eccentricity and Directrix
An alternative way to describe a conic section involves the directrices, the foci, and a new property called eccentricity. We will see that the value of the eccentricity of a conic section can uniquely define that conic.
The three conic sections with their directrices appear in the following figure.
Recall from the definition of a parabola that the distance from any point on the parabola to the focus is equal to the distance from that same point to the directrix. Therefore, by definition, the eccentricity of a parabola must be 1. The equations of the directrices of a horizontal ellipse are \(x=\text{\pm }\frac{{a}^{2}}{c}.\) The right vertex of the ellipse is located at \((a,0)\) and the right focus is \((c,0).\) Therefore the distance from the vertex to the focus is \(a-c\) and the distance from the vertex to the right directrix is \(\frac{{a}^{2}}{c}-a.\) This gives the eccentricity as
\[e=\frac{a-c}{\frac{{a}^{2}}{c}-a}=\frac{c(a-c)}{{a}^{2}-ac}=\frac{c(a-c)}{a(a-c)}=\frac{c}{a}.\]Since \(c
Example
Try it.
Determine the eccentricity of the ellipse described by the equation
\[\frac{{(x-3)}^{2}}{16}+\frac{{(y+2)}^{2}}{25}=1.\]Solution
From the equation we see that \(a=5\) and \(b=4.\) The value of c can be calculated using the equation \({a}^{2}={b}^{2}+{c}^{2}\) for an ellipse. Substituting the values of a and b and solving for c gives \(c=3.\) Therefore the eccentricity of the ellipse is \(e=\frac{c}{a}=\frac{3}{5}=0.6.\)
Polar Equations of Conic Sections
Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter. The focal parameter of a conic section p is defined as the distance from a focus to the nearest directrix. The following table gives the focal parameters for the different types of conics, where a is the length of the semi-major axis (i.e., half the length of the major axis), c is the distance from the origin to the focus, and e is the eccentricity. In the case of a parabola, a represents the distance from the vertex to the focus.
| Conic | e | p |
| Ellipse | \(0| \(\frac{{a}^{2}-{c}^{2}}{c}=\frac{a(1-{e}^{2})}{e}\) | |
| Parabola | \(e=1\) | \(2a\) |
| Hyperbola | \(e>1\) | \(\frac{{c}^{2}-{a}^{2}}{c}=\frac{a({e}^{2}-1)}{e}\) |
Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conic section in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then using the definition of the various conic sections in terms of distances, it is possible to prove the following theorem.
In the equation on the left, the major axis of the conic section is horizontal, and in the equation on the right, the major axis is vertical. To work with a conic section written in polar form, first make the constant term in the denominator equal to 1. This can be done by dividing both the numerator and the denominator of the fraction by the constant that appears in front of the plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. This value identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conic is vertical. If both appear then the axes are rotated. The center of the conic is not necessarily at the origin. The center is at the origin only if the conic is a circle (i.e., \(e=0).\)
Condensed — the full section is in OpenStax Calculus Volume 2.
General Equations of Degree Two
A general equation of degree two can be written in the form
\[A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0.\]The graph of an equation of this form is a conic section. If \(B\ne 0\) then the coordinate axes are rotated. To identify the conic section, we use the discriminant of the conic section \(4AC-{B}^{2}.\) One of the following cases must be true:
- \(4AC-{B}^{2}>0.\) If so, the graph is an ellipse.
- \(4AC-{B}^{2}=0.\) If so, the graph is a parabola.
- \(4AC-{B}^{2}<0.\) If so, the graph is a hyperbola.
The simplest example of a second-degree equation involving a cross term is \(xy=1.\) This equation can be solved for y to obtain \(y=\frac{1}{x}.\) The graph of this function is called a rectangular hyperbola as shown.
The asymptotes of this hyperbola are the x and y coordinate axes. To determine the angle \(\theta\) of rotation of the conic section, we use the formula \(\text{cot}\ 2\theta =\frac{A-C}{B}.\) In this case \(A=C=0\) and \(B=1,\) so \(\text{cot}\ 2\theta =(0-0)\text{/}1=0\) and \(\theta =45\text{^{\circ}}.\) The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled \({A}^{'},{B}^{'},{C}^{'},{D}^{'},{E}^{'},\ \text{and}\ {F}^{'},\) and are given by the formulas
\[\begin{array}{lll}{A}^{'} & = & A\ {\text{cos}}^{2}\theta +B\ \text{cos}\ \theta \ \text{sin}\ \theta +C\ {\text{sin}}^{2}\theta \\ {B}^{'} & = & 0 \\ {C}^{'} & = & A\ {\text{sin}}^{2}\theta -B\ \text{sin}\ \theta \ \text{cos}\ \theta +C\ {\text{cos}}^{2}\theta \\ {D}^{'} & = & D\ \text{cos}\ \theta +E\ \text{sin}\ \theta \\ {E}^{'} & = & \text{-}D\ \text{sin}\ \theta +E\ \text{cos}\ \theta \\ {F}^{'} & = & F.\end{array}\]The procedure for graphing a rotated conic is the following:
- Identify the conic section using the discriminant \(4AC-{B}^{2}.\)
- Determine \(\theta\) using the formula \(\text{cot}\ 2\theta =\frac{A-C}{B}.\)
- Calculate \({A}^{'},{B}^{'},{C}^{'},{D}^{'},{E}^{'},\ \text{and}\ {F}^{'}.\)
- Rewrite the original equation using \({A}^{'},{B}^{'},{C}^{'},{D}^{'},{E}^{'},\ \text{and}\ {F}^{'}.\)
- Draw a graph using the rotated equation.
Condensed — the full section is in OpenStax Calculus Volume 2.
Key Concepts
- The equation of a vertical parabola in standard form with given focus and directrix is \(y=\frac{1}{4p}{(x-h)}^{2}+k\) where p is the distance from the vertex to the focus and \((h,k)\) are the coordinates of the vertex.
- The equation of a horizontal ellipse in standard form is \(\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1\) where the center has coordinates \((h,k),\) the major axis has length 2a, the minor axis has length 2b, and the coordinates of the foci are \((h\pm c,k),\) where \({c}^{2}={a}^{2}-{b}^{2}.\)
- The equation of a horizontal hyperbola in standard form is \(\frac{{(x-h)}^{2}}{{a}^{2}}-\frac{{(y-k)}^{2}}{{b}^{2}}=1\) where the center has coordinates \((h,k),\) the vertices are located at \((h\pm a,k),\) and the coordinates of the foci are \((h\pm c,k),\) where \({c}^{2}={a}^{2}+{b}^{2}.\)
- The eccentricity of an ellipse is less than 1, the eccentricity of a parabola is equal to 1, and the eccentricity of a hyperbola is greater than 1. The eccentricity of a circle is 0.
- The polar equation of a conic section with eccentricity e is \(r=\frac{ep}{1\pm e\ \text{cos}\ \theta }\) or \(r=\frac{ep}{1\pm e\ \text{sin}\ \theta },\) where p represents the focal parameter.
- To identify a conic generated by the equation \(A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0,\) first calculate the discriminant \(D=4AC-{B}^{2}.\) If \(D>0\) then the conic is an ellipse, if \(D=0\) then the conic is a parabola, and if \(D<0\) then the conic is a hyperbola.
Conic Sections
For the following exercises, determine the equation of the parabola using the information given.
For the following exercises, determine the equation of the ellipse using the information given.
For the following exercises, determine the equation of the hyperbola using the information given.
For the following exercises, consider the following polar equations of conics. Determine the eccentricity and identify the conic.
For the following exercises, find a polar equation of the conic with focus at the origin and eccentricity and directrix as given.
For the following exercises, sketch the graph of each conic.
For the following equations, determine which of the conic sections is described.
Condensed — the full section is in OpenStax Calculus Volume 2.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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Put the equation \({x}^{2}-4x-8y+12=0\) into standard form and graph the resulting parabola.
Odkrij odgovor
Since y is not squared in this equation, we know that the parabola opens either upward or downward. Therefore we need to solve this equation for y, which will put the equation into standard form. To do that, first add \(8y\) to both sides of the equation:
\[8y={x}^{2}-4x+12.\]The next step is to complete the square on the right-hand side. Start by grouping the first two terms on the right-hand side using parentheses:
\[8y=({x}^{2}-4x)+12.\]Next determine the constant that, when added inside the parentheses, makes the quantity inside the parentheses a perfect square trinomial. To do this, take half the coefficient of x and square it. This gives \({(\frac{-4}{2})}^{2}=4.\) Add 4 inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:
\[8y=({x}^{2}-4x+4)+12-4.\]Now combine like terms and factor the quantity inside the parentheses:
\[8y={(x-2)}^{2}+8.\]Finally, divide by 8:
\[y=\frac{1}{8}{(x-2)}^{2}+1.\]This equation is now in standard form. Comparing this to gives \(h=2,\) \(k=1,\) and \(p=2.\) The parabola opens up, with vertex at \((2,1),\) focus at \((2,3),\) and directrix \(y=-1.\) The graph of this parabola appears as follows.
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Put the equation \(2{y}^{2}-x+12y+16=0\) into standard form and graph the resulting parabola.
Odkrij odgovor
\(x=2{(y+3)}^{2}-2\)
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Put the equation \(9{x}^{2}+4{y}^{2}-36x+24y+36=0\) into standard form and graph the resulting ellipse.
Odkrij odgovor
First subtract 36 from both sides of the equation:
\[9{x}^{2}+4{y}^{2}-36x+24y=-36.\]Next group the x terms together and the y terms together, and factor out the common factor:
\[\begin{array}{lll}(9{x}^{2}-36x)+(4{y}^{2}+24y) & = & -36 \\ 9({x}^{2}-4x)+4({y}^{2}+6y) & = & -36.\end{array}\]We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives \({(\frac{-4}{2})}^{2}=4.\) In the second set of parentheses, take half the coefficient of y and square it. This gives \({(\frac{6}{2})}^{2}=9.\) Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are adding 36 to the second set as well. Therefore the equation becomes
\[\begin{array}{l} \\ 9({x}^{2}-4x+4)+4({y}^{2}+6y+9)=-36+36+36 \\ 9({x}^{2}-4x+4)+4({y}^{2}+6y+9)=36.\end{array}\]Now factor both sets of parentheses and divide by 36:
\[\begin{array}{lll} \\ 9{(x-2)}^{2}+4{(y+3)}^{2} & = & 36 \\ \frac{9{(x-2)}^{2}}{36}+\frac{4{(y+3)}^{2}}{36} & = & 1 \\ \frac{{(x-2)}^{2}}{4}+\frac{{(y+3)}^{2}}{9} & = & 1.\end{array}\]The equation is now in standard form. Comparing this to gives \(h=2,\) \(k=-3,\) \(a=3,\) and \(b=2.\) This is a vertical ellipse with center at \((2,-3),\) major axis 6, and minor axis 4. The graph of this ellipse appears as follows.
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Put the equation \(9{x}^{2}+16{y}^{2}+18x-64y-71=0\) into standard form and graph the resulting ellipse.
Odkrij odgovor
\(\frac{{(x+1)}^{2}}{16}+\frac{{(y-2)}^{2}}{9}=1\)
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Put the equation \(9{x}^{2}-16{y}^{2}+36x+32y-124=0\) into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?
Odkrij odgovor
First add 124 to both sides of the equation:
\[9{x}^{2}-16{y}^{2}+36x+32y=124.\]Next group the x terms together and the y terms together, then factor out the common factors:
\[\begin{array}{lll}(9{x}^{2}+36x)-(16{y}^{2}-32y) & = & 124 \\ 9({x}^{2}+4x)-16({y}^{2}-2y) & = & 124.\end{array}\]We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives \({(\frac{4}{2})}^{2}=4.\) In the second set of parentheses, take half the coefficient of y and square it. This gives \({(\frac{-2}{2})}^{2}=1.\) Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes
\[\begin{array}{l} \\ 9({x}^{2}+4x+4)-16({y}^{2}-2y+1)=124+36-16 \\ 9({x}^{2}+4x+4)-16({y}^{2}-2y+1)=144.\end{array}\]Next factor both sets of parentheses and divide by 144:
\[\begin{array}{lll} \\ 9{(x+2)}^{2}-16{(y-1)}^{2} & = & 144 \\ \frac{9{(x+2)}^{2}}{144}-\frac{16{(y-1)}^{2}}{144} & = & 1 \\ \frac{{(x+2)}^{2}}{16}-\frac{{(y-1)}^{2}}{9} & = & 1.\end{array}\]The equation is now in standard form. Comparing this to gives \(h=-2,\) \(k=1,\) \(a=4,\) and \(b=3.\) This is a horizontal hyperbola with center at \((-2,1)\) and asymptotes given by the equations \(y=1\pm \frac{3}{4}(x+2).\) The graph of this hyperbola appears in the following figure.
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Put the equation \(4{y}^{2}-9{x}^{2}+16y+18x-29=0\) into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?
Odkrij odgovor
\(\frac{{(y+2)}^{2}}{9}-\frac{{(x-1)}^{2}}{4}=1.\) This is a vertical hyperbola. Asymptotes \(y=-2\pm \frac{3}{2}(x-1).\)
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Determine the eccentricity of the ellipse described by the equation
\[\frac{{(x-3)}^{2}}{16}+\frac{{(y+2)}^{2}}{25}=1.\]Odkrij odgovor
From the equation we see that \(a=5\) and \(b=4.\) The value of c can be calculated using the equation \({a}^{2}={b}^{2}+{c}^{2}\) for an ellipse. Substituting the values of a and b and solving for c gives \(c=3.\) Therefore the eccentricity of the ellipse is \(e=\frac{c}{a}=\frac{3}{5}=0.6.\)
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Determine the eccentricity of the hyperbola described by the equation
\[\frac{{(y-3)}^{2}}{49}-\frac{{(x+2)}^{2}}{25}=1.\]Odkrij odgovor
\(e=\frac{c}{a}=\frac{\sqrt{74}}{7}\approx 1.229\)
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Identify and create a graph of the conic section described by the equation
\[r=\frac{3}{1+2\ \text{cos}\ \theta }.\]Odkrij odgovor
The constant term in the denominator is 1, so the eccentricity of the conic is 2. This is a hyperbola. The focal parameter p can be calculated by using the equation \(ep=3.\) Since \(e=2,\) this gives \(p=\frac{3}{2}.\) The cosine function appears in the denominator, so the hyperbola is horizontal. Pick a few values for \(\theta\) and create a table of values. Then we can graph the hyperbola ().
\(\theta\) \(r\) \(\theta\) \(r\) 0 1 \(\pi\) −3 \(\frac{\pi }{4}\) \(\frac{3}{1+\sqrt{2}}\approx 1.2426\) \(\frac{5\pi }{4}\) \(\frac{3}{1-\sqrt{2}}\approx -7.2426\) \(\frac{\pi }{2}\) 3 \(\frac{3\pi }{2}\) 3 \(\frac{3\pi }{4}\) \(\frac{3}{1-\sqrt{2}}\approx -7.2426\) \(\frac{7\pi }{4}\) \(\frac{3}{1+\sqrt{2}}\approx 1.2426\) -
Identify and create a graph of the conic section described by the equation
\[r=\frac{4}{1-0.8\ \text{sin}\ \theta }.\]Odkrij odgovor
Here \(e=0.8\) and \(p=5.\) This conic section is an ellipse.
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Identify the conic and calculate the angle of rotation of axes for the curve described by the equation
\[13{x}^{2}-6\sqrt{3}xy+7{y}^{2}-256=0.\]Odkrij odgovor
In this equation, \(A=13,B=-6\sqrt{3},C=7,D=0,E=0,\) and \(F=-256.\) The discriminant of this equation is \(4AC-{B}^{2}=4(13)(7)-{(-6\sqrt{3})}^{2}=364-108=256.\) Therefore this conic is an ellipse. To calculate the angle of rotation of the axes, use \(\text{cot}\ 2\theta =\frac{A-C}{B}.\) This gives
\[\begin{array}{ll}\text{cot}\ 2\theta & =\frac{A-C}{B} \\ & =\frac{13-7}{-6\sqrt{3}} \\ & =-\frac{\sqrt{3}}{3}.\end{array}\]Therefore \(2\theta ={120}^{\text{o}}\) and \(\theta ={60}^{\text{o}},\) which is the angle of the rotation of the axes.
To determine the rotated coefficients, use the formulas given above:
\[\begin{array}{lll}{A}^{'} & = & A\ {\text{cos}}^{2}\theta +B\ \text{cos}\ \theta \ \text{sin}\ \theta +C\ {\text{sin}}^{2}\theta \\ & = & 13{\text{cos}}^{2}60+(-6\sqrt{3})\ \text{cos}\ 60\ \text{sin}\ 60+7{\text{sin}}^{2}60 \\ & = & 13{(\frac{1}{2})}^{2}-6\sqrt{3}(\frac{1}{2})(\frac{\sqrt{3}}{2})+7{(\frac{\sqrt{3}}{2})}^{2} \\ & = & 4, \\ {B}^{'} & = & 0, \\ {C}^{'} & = & A\ {\text{sin}}^{2}\theta -B\ \text{sin}\ \theta \ \text{cos}\ \theta +C\ {\text{cos}}^{2}\theta \\ & = & 13{\text{sin}}^{2}60+(-6\sqrt{3})\ \text{sin}\ 60\ \text{cos}\ 60=7{\text{cos}}^{2}60 \\ & = & {(\frac{\sqrt{3}}{2})}^{2}+6\sqrt{3}(\frac{\sqrt{3}}{2})(\frac{1}{2})+7{(\frac{1}{2})}^{2} \\ & = & 16, \\ {D}^{'} & = & D\ \text{cos}\ \theta +E\ \text{sin}\ \theta \\ & = & (0)\ \text{cos}\ 60+(0)\ \text{sin}\ 60 \\ & = & 0, \\ {E}^{'} & = & \text{-}D\ \text{sin}\ \theta +E\ \text{cos}\ \theta \\ & = & \text{-}(0)\ \text{sin}\ 60+(0)\ \text{cos}\ 60 \\ & = & 0, \\ {F}^{'} & = & F \\ & = & -256.\end{array}\]The equation of the conic in the rotated coordinate system becomes
\[\begin{array}{lll} \\ 4{({x}^{'})}^{2}+16{({y}^{'})}^{2} & = & 256 \\ \frac{{({x}^{'})}^{2}}{64}+\frac{{({y}^{'})}^{2}}{16} & = & 1.\end{array}\]A graph of this conic section appears as follows.
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Identify the conic and calculate the angle of rotation of axes for the curve described by the equation
\[3{x}^{2}+5xy-2{y}^{2}-125=0.\]Odkrij odgovor
The conic is a hyperbola and the angle of rotation of the axes is \(\theta =22.5\text{^{\circ}}.\)
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Focus \((4,0)\) and directrix \(x=-4\)
Odkrij odgovor
\({y}^{2}=16x\)
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Focus \((0,-3)\) and directrix \(y=3\)
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Focus \((0,0.5)\) and directrix \(y=-0.5\)
Odkrij odgovor
\({x}^{2}=2y\)
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Focus \((2,\ 3)\) and directrix \(x=-2\)
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Focus \((0,2)\) and directrix \(y=4\)
Odkrij odgovor
\({x}^{2}=-4(y-3)\)
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Focus \((-1,4)\) and directrix \(x=5\)
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Focus \((-3,5)\) and directrix \(y=1\)
Odkrij odgovor
\({(x+3)}^{2}=8(y-3)\)
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Focus \((\frac{5}{2},-4)\) and directrix \(x=\frac{7}{2}\)
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Endpoints of major axis at \((4,0),(-4,0)\) and foci located at \((2,0),(-2,0)\)
Odkrij odgovor
\(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{12}=1\)
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Endpoints of major axis at \((0,5),(0,-5)\) and foci located at \((0,3),(0,-3)\)
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Endpoints of minor axis at \((0,2),(0,-2)\) and foci located at \((3,0),(-3,0)\)
Odkrij odgovor
\(\frac{{x}^{2}}{13}+\frac{{y}^{2}}{4}=1\)
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Endpoints of major axis at \((-3,3),(7,3)\) and foci located at \((-2,3),(6,3)\)
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Endpoints of major axis at \((-3,5),(-3,-3)\) and foci located at \((-3,3),(-3,-1)\)
Odkrij odgovor
\(\frac{{(y-1)}^{2}}{16}+\frac{{(x+3)}^{2}}{12}=1\)
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Endpoints of minor axis at \((0,0),(0,4)\) and foci located at \((5,2),(-5,2)\)
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Foci located at \((2,0),\ (-2,0)\) and eccentricity of \(\frac{1}{2}\)
Odkrij odgovor
\(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{12}=1\)
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Foci located at \((0,-3),\ (0,3)\) and eccentricity of \(\frac{3}{4}\)
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Vertices located at \((5,0),(-5,0)\) and foci located at \((6,0),(-6,0)\)
Odkrij odgovor
\(\frac{{x}^{2}}{25}-\frac{{y}^{2}}{11}=1\)
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Vertices located at \((0,2),(0,-2)\) and foci located at \((0,3),(0,-3)\)
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Endpoints of the conjugate axis located at \((0,3),(0,-3)\) and foci located \((4,0),(-4,0)\)
Odkrij odgovor
\(\frac{{x}^{2}}{7}-\frac{{y}^{2}}{9}=1\)
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Vertices located at \((0,1),(6,1)\) and focus located at \((8,1)\)
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Vertices located at \((-2,0),(-2,-4)\) and focus located at \((-2,-8)\)
Odkrij odgovor
\(\frac{{(y+2)}^{2}}{4}-\frac{{(x+2)}^{2}}{32}=1\)
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Endpoints of the conjugate axis located at \((3,2),(3,4)\) and focus located at \((3,7)\)
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Foci located at \((-6,0),(6,0)\) and eccentricity of 3
Odkrij odgovor
\(\frac{{x}^{2}}{4}-\frac{{y}^{2}}{32}=1\)
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\((0,10),(0,-10)\) and eccentricity of 2.5
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\(r=\frac{-1}{1+\text{cos}\ \theta }\)
Odkrij odgovor
\(e=1,\) parabola
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\(r=\frac{8}{2-\text{sin}\ \theta }\)
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\(r=\frac{5}{2+\text{sin}\ \theta }\)
Odkrij odgovor
\(e=\frac{1}{2},\) ellipse
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\(r=\frac{5}{-1+2\ \text{sin}\ \theta }\)
Symbols used here
The non-negative number whose square (n-th power) is x.
The usual name for an angle.
i² = −1.
1/360 of a full turn. 180° = π radians.
Equal to the precision shown, not exactly.
Both signs at once: x = 3 ± 2 means 5 and 1.
The two sides are different.
Derivative with respect to x, holding the other variables fixed.
Vector of partial derivatives; points uphill.
Antiderivative (indefinite) or signed area from a to b (definite).
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: Conic Sections
- Identify the equation of a parabola in standard form with given focus and directrix.
- Identify the equation of an ellipse in standard form with given foci.
- Identify the equation of a hyperbola in standard form with given foci.
- Recognize a parabola, ellipse, or hyperbola from its eccentricity value.
- Write the polar equation of a conic section with eccentricity
- Identify when a general equation of degree two is a parabola, ellipse, or hyperbola.
- If
- If
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
Poskusi sam.
Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
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Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems