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Ellipses

Graph an ellipse with center at the origin

Graph an Ellipse with Center at the Origin

The next conic section we will look at is an ellipse. We define an ellipse as all points in a plane where the sum of the distances from two fixed points is constant. Each of the given points is called a focus of the ellipse.

We can draw an ellipse by taking some fixed length of flexible string and attaching the ends to two thumbtacks. We use a pen to pull the string taut and rotate it around the two thumbtacks. The figure that results is an ellipse.

A line drawn through the foci intersect the ellipse in two points. Each point is called a vertex of the ellipse. The segment connecting the vertices is called the major axis. The midpoint of the segment is called the center of the ellipse. A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the minor axis.

We mentioned earlier that our goal is to connect the geometry of a conic with algebra. Placing the ellipse on a rectangular coordinate system gives us that opportunity. In the figure, we placed the ellipse so the foci \(((\text{-}c,0),(c,0))\) are on the x-axis and the center is the origin.

The definition states the sum of the distance from the foci to a point \((x,y)\) is constant. So \({d}_{1}+{d}_{2}\) is a constant that we will call \(2a\) so, \({d}_{1}+{d}_{2}=2a.\) We will use the distance formula to lead us to an algebraic formula for an ellipse.

\[\begin{array}{llllllllllll}\begin{array}{l} \\ \\ \\ \\ \\ \text{Use the distance formula to find}\ {d}_{1},{d}_{2}.\end{array} & & & \begin{array}{lllll}{d}_{1} & + & {d}_{2} & = & 2a \\ \\ \\ \sqrt{{(x-(\text{-}c))}^{2}+{(y-0)}^{2}} & + & \sqrt{{(x-c)}^{2}+{(y-0)}^{2}} & = & 2a\end{array} \\ \\ \\ \begin{array}{l}\text{After eliminating radicals and simplifying,} \\ \text{we get:}\end{array} & & & \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{a}^{2}-{c}^{2}}\ =\ 1\ \\ \begin{array}{l}\text{To simplify the equation of the ellipse, we} \\ \text{let}\ {a}^{2}-{c}^{2}={b}^{2}.\end{array} & & & \\ \begin{array}{l}\text{So, the equation of an ellipse centered at the} \\ \text{origin in standard form is:}\end{array} & & & \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}\ =\ 1\ \end{array}\]

To graph the ellipse, it will be helpful to know the intercepts. We will find the x-intercepts and y-intercepts using the formula.

How to Graph an Ellipse with Center (0, 0)

Try it.

Graph: \(\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1.\)

Solution

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Find the Equation of an Ellipse with Center at the Origin

If we are given the graph of an ellipse, we can find the equation of the ellipse.

Example

Try it.

Find the equation of the ellipse shown.

Solution

We recognize this as an ellipse that is centered at the origin.\(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\)
Since the major axis is horizontal and the distance from the center to the vertex is 4, we know \(a=4\) and so \({a}^{2}=16\).\(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{{b}^{2}}=1\)
The minor axis is vertical and the distance from the center to the ellipse is 3, we know \(b=3\) and so \({b}^{2}=9\).\(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1\)

Graph an Ellipse with Center Not at the Origin

The ellipses we have looked at so far have all been centered at the origin. We will now look at ellipses whose center is \((h,k).\)

The equation is \(\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1\) and when \(a>b,\) the major axis is horizontal so the distance from the center to the vertex is a. When \(b>a,\) the major axis is vertical so the distance from the center to the vertex is b.

Example

Try it.

Graph: \(\frac{{(x-3)}^{2}}{9}+\frac{{(y-1)}^{2}}{4}=1.\)

Solution

The equation is in standard form,
\(\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1.\)		\(\frac{{(x-3)}^{2}}{9}+\frac{{(y-1)}^{2}}{4}=1\)
The ellipse is centered at \((h,k).\)The center is \((3,1).\)
Since \(9>4\) and 9 is in the \({x}^{2}\) term,
the major axis is horizontal.
  \({a}^{2}=9,\ a=\pm 3\)
  \({b}^{2}=4,\ b=\pm 2\)		The distance from the center to the vertices is 3.
The distance from the center to the endpoints of the
minor axis is 2.
Sketch the ellipse.

If we look at the equations of \(\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1\) and \(\frac{{(x-3)}^{2}}{9}+\frac{{(y-1)}^{2}}{4}=1,\) we see that they are both ellipses with \(a=3\) and \(b=2.\) So they will have the same size and shape. They are different in that they do not have the same center.

Notice in the graph above that we could have graphed \(\frac{{(x-3)}^{2}}{9}+\frac{{(y-1)}^{2}}{4}=1\) by translations. We moved the original ellipse to the right 3 units and then up 1 unit.

In the next example we will use the translation method to graph the ellipse.

When an equation has both an \({x}^{2}\) and a \({y}^{2}\) with different coefficients, we verify that it is an ellipsis by putting it in standard form. We will then be able to graph the equation.

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

  • Ellipse: An ellipse is all points in a plane where the sum of the distances from two fixed points is constant. Each of the fixed points is called a focus of the ellipse.

    If we draw a line through the foci intersects the ellipse in two points, each is called a vertex of the ellipse.
    The segment connecting the vertices is called the major axis.
    The midpoint of the segment is called the center of the ellipse.
    A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the minor axis.
  • Standard Form of the Equation an Ellipse with Center \((0,0):\) The standard form of the equation of an ellipse with center \((0,0),\) is
    \[\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\]
    The x-intercepts are \((a,0)\) and \((\text{-}a,0).\)
    The y-intercepts are \((0,b)\) and \((0,\text{-}b).\)
  • How to an Ellipse with Center \((0,0)\)
    1. Write the equation in standard form.
    2. Determine whether the major axis is horizontal or vertical.
    3. Find the endpoints of the major axis.
    4. Find the endpoints of the minor axis
    5. Sketch the ellipse.
  • Standard Form of the Equation an Ellipse with Center \((h,k):\) The standard form of the equation of an ellipse with center \((h,k),\) is
    \[\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1\]
    When \(a>b,\) the major axis is horizontal so the distance from the center to the vertex is a.
    When \(b>a,\) the major axis is vertical so the distance from the center to the vertex is b.

Ellipses

Graph an Ellipse with Center at the Origin

In the following exercises, graph each ellipse.

Try it.

\(\frac{{x}^{2}}{4}+\frac{{y}^{2}}{25}=1\)

Solution

Try it.

\(\frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1\)

Try it.

\(\frac{{x}^{2}}{25}+\frac{{y}^{2}}{36}=1\)

Solution

Try it.

\(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{36}=1\)

Try it.

\(\frac{{x}^{2}}{36}+\frac{{y}^{2}}{16}=1\)

Solution

Try it.

\(\frac{{x}^{2}}{25}+\frac{{y}^{2}}{9}=1\)

Try it.

\({x}^{2}+\frac{{y}^{2}}{4}=1\)

Solution

Try it.

\(\frac{{x}^{2}}{9}+{y}^{2}=1\)

Try it.

\(4{x}^{2}+25{y}^{2}=100\)

Solution

Try it.

\(16{x}^{2}+9{y}^{2}=144\)

Try it.

\(16{x}^{2}+36{y}^{2}=576\)

Solution

Try it.

\(9{x}^{2}+25{y}^{2}=225\)

Find the Equation of an Ellipse with Center at the Origin

In the following exercises, find the equation of the ellipse shown in the graph.

Graph an Ellipse with Center Not at the Origin

In the following exercises, graph each ellipse.

Try it.

\(\frac{{(x+1)}^{2}}{4}+\frac{{(y+6)}^{2}}{25}=1\)

Solution

Try it.

\(\frac{{(x-3)}^{2}}{25}+\frac{{(y+2)}^{2}}{9}=1\)

Try it.

\(\frac{{(x+4)}^{2}}{4}+\frac{{(y-2)}^{2}}{9}=1\)

Solution

Try it.

\(\frac{{(x-4)}^{2}}{9}+\frac{{(y-1)}^{2}}{16}=1\)

In the following exercises, graph each equation by translation.

Try it.

\(\frac{{(x-3)}^{2}}{4}+\frac{{(y-7)}^{2}}{25}=1\)

Solution

Try it.

\(\frac{{(x+6)}^{2}}{16}+\frac{{(y+5)}^{2}}{4}=1\)

Try it.

\(\frac{{(x-5)}^{2}}{9}+\frac{{(y+4)}^{2}}{25}=1\)

Solution

Try it.

\(\frac{{(x+5)}^{2}}{36}+\frac{{(y-3)}^{2}}{16}=1\)

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

Try it.

\(25{x}^{2}+9{y}^{2}-100x-54y-44=0\)

Solution

ⓐ \(\frac{{(x-2)}^{2}}{9}+\frac{{(y-3)}^{2}}{25}=1\)
ⓑ

Try it.

\(4{x}^{2}+25{y}^{2}+8x+100y+4=0\)

Try it.

\(4{x}^{2}+25{y}^{2}-24x-64=0\)

Solution

ⓐ \(\frac{{y}^{2}}{4}+\frac{{(x-3)}^{2}}{25}=1\)
ⓑ

Try it.

\(9{x}^{2}+4{y}^{2}+56y+160=0\)

In the following exercises, graph the equation.

Try it.

\(x=-2{(y-1)}^{2}+2\)

Solution

Try it.

\({x}^{2}+{y}^{2}=49\)

Try it.

\({(x+5)}^{2}+{(y+2)}^{2}=4\)

Solution

Try it.

\(y=\text{-}{x}^{2}+8x-15\)

Try it.

\(\frac{{(x+3)}^{2}}{16}+\frac{{(y+1)}^{2}}{4}=1\)

Solution

Try it.

\({(x-2)}^{2}+{(y-3)}^{2}=9\)

Try it.

\(\frac{{x}^{2}}{25}+\frac{{y}^{2}}{36}=1\)

Solution

Try it.

\(x=4{(y+1)}^{2}-4\)

Try it.

\({x}^{2}+{y}^{2}=64\)

Solution

Try it.

\(\frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1\)

Try it.

\(y=6{x}^{2}+2x-1\)

Solution

Try it.

\(\frac{{(x-2)}^{2}}{9}+\frac{{(y+3)}^{2}}{25}=1\)

Solve Application with Ellipses

Try it.

A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 10 AU and the furthest is approximately 30 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

Solution

\(\frac{{x}^{2}}{400}+\frac{{y}^{2}}{300}=1\)

Try it.

A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 10 AU and the furthest is approximately 70 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

Try it.

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 15 AU and the furthest is approximately 85 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

Solution

\(\frac{{x}^{2}}{2500}+\frac{{y}^{2}}{1275}=1\)

Try it.

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 15 AU and the furthest is approximately 95 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

이제 너 계산기는 이것을 해결하지 않지만, 그 조각은 계산 가능합니다. 아래의 하나를 시도하거나 자신의 것을 입력하십시오.

연습 (40)

먼저 종이에 각각을 시도해 보세요. 확인하기 위해 답을 드러내세요. 확인된 답은 각 단계의 솔버에서 열 수 있습니다.

  1. Graph \(y={(x-1)}^{2}-2\) using transformations.

  2. Complete the square: \({x}^{2}-8x=8.\)

    답을 드러내세요

    \({\left(x-4\right)}^{2}=8\)

  3. Write in standard form. \(y=2{x}^{2}-12x+14\)

    답을 드러내세요

    \(y=2{\left(x-3\right)}^{2}-4\)

  4. Graph: \(\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1.\)

  5. Graph: \(\frac{{x}^{2}}{4}+\frac{{y}^{2}}{16}=1.\)

  6. Graph: \(\frac{{x}^{2}}{9}+\frac{{y}^{2}}{16}=1.\)

  7. Graph \({x}^{2}+4{y}^{2}=16.\)

    답을 드러내세요

    We recognize this as the equation of an
    ellipse since both the x and y terms are
    squared and have different coefficients.    		\(\ {x}^{2}+4{y}^{2}=16\)
    To get the equation in standard form, divide
    both sides by 16 so that the equation is equal
    to 1.    		\(\ \frac{{x}^{2}}{16}+\frac{4{y}^{2}}{16}=\frac{16}{16}\)
    Simplify.\(\ \frac{{x}^{2}}{16}+\frac{{y}^{2}}{4}=1\)
    The equation is in standard form.
    The ellipse is centered at the origin.    		The center is \((0,0).\)
    Since \(16>4\) and 16 is in the \({x}^{2}\) term,
    the major axis is horizontal.
      \({a}^{2}=16,a=\pm 4\)
      \({b}^{2}=4,\ b=\pm 2\)    		The vertices are \((4,0),(-4,0).\)
    The endpoints of the minor axis are
    \((0,2),(0,-2).\)
    Sketch the ellipse.

  8. Graph \(9{x}^{2}+16{y}^{2}=144.\)

  9. Graph \(16{x}^{2}+25{y}^{2}=400.\)

  10. Find the equation of the ellipse shown.

    답을 드러내세요

    We recognize this as an ellipse that is centered at the origin.\(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\)
    Since the major axis is horizontal and the distance from the center to the vertex is 4, we know \(a=4\) and so \({a}^{2}=16\).\(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{{b}^{2}}=1\)
    The minor axis is vertical and the distance from the center to the ellipse is 3, we know \(b=3\) and so \({b}^{2}=9\).\(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1\)

  11. Find the equation of the ellipse shown.

    답을 드러내세요

    \(\frac{{x}^{2}}{4}+\frac{{y}^{2}}{25}=1\)

  12. Find the equation of the ellipse shown.

    답을 드러내세요

    \(\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1\)

  13. Graph: \(\frac{{(x-3)}^{2}}{9}+\frac{{(y-1)}^{2}}{4}=1.\)

    답을 드러내세요

    The equation is in standard form,
    \(\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1.\)    		\(\frac{{(x-3)}^{2}}{9}+\frac{{(y-1)}^{2}}{4}=1\)
    The ellipse is centered at \((h,k).\)The center is \((3,1).\)
    Since \(9>4\) and 9 is in the \({x}^{2}\) term,
    the major axis is horizontal.
      \({a}^{2}=9,\ a=\pm 3\)
      \({b}^{2}=4,\ b=\pm 2\)    		The distance from the center to the vertices is 3.
    The distance from the center to the endpoints of the
    minor axis is 2.
    Sketch the ellipse.

  14. Graph: \(\frac{{(x+3)}^{2}}{4}+\frac{{(y-5)}^{2}}{16}=1.\)

  15. Graph: \(\frac{{(x-1)}^{2}}{25}+\frac{{(y+3)}^{2}}{16}=1.\)

  16. Graph \(\frac{{(x+4)}^{2}}{16}+\frac{{(y-6)}^{2}}{9}=1\) by translation.

    답을 드러내세요

    This ellipse will have the same size and shape as \(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1\) whose center is \((0,0).\) We graph this ellipse first.

    The center is \((0,0).\)Center \((0,0)\)
    Since \(16>9,\) the major axis is horizontal.
      \({a}^{2}=16,a=\pm 4\)
      \({b}^{2}=9,\ b=\pm 3\)    		The vertices are \((4,0),(-4,0).\)
    The endpoints of the minor axis are
    \((0,3),(0,-3).\)
    Sketch the ellipse.
    The original equation is in standard form,
    \(\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1.\)    		\(\frac{{(x-(-4))}^{2}}{16}+\frac{{(y-6)}^{2}}{9}=1\)
    The ellipse is centered at \((h,k).\)The center is \((-4,6).\)
    We translate the graph of \(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1\) four
    units to the left and then up 6 units.
    Verify that the center is \((-4,6).\)
    The new ellipse is the ellipse whose equation
    is
    \(\frac{{(x+4)}^{2}}{16}+\frac{{(y-6)}^{2}}{9}=1.\)

  17. Graph \(\frac{{(x-5)}^{2}}{9}+\frac{{(y+4)}^{2}}{4}=1\) by translation.

  18. Graph \(\frac{{(x+6)}^{2}}{16}+\frac{{(y+2)}^{2}}{25}=1\) by translation.

  19. Write the equation \({x}^{2}+4{y}^{2}-4x+24y+24=0\) in standard form and graph.

    답을 드러내세요

    We put the equation in standard form by completing the squares in both x and y.

    \({x}^{2}+4{y}^{2}-4x+24y+24=0\)
    Rewrite grouping the x terms and y terms.
    Make the coefficients of \({x}^{2}\) and \({y}^{2}\) equal 1.
    Complete the squares.
    Write as binomial squares.
    Divide both sides by 16 to get 1 on the right.
    Simplify.
    The equation is in standard form,
    \(\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1\)
    The ellipse is centered at \((h,k).\)The center is \((2,-3).\)
    Since \(16>4\) and 16 is in the \({x}^{2}\) term,
    the major axis is horizontal.
      \({a}^{2}=16,a=\pm 4\)
      \({b}^{2}=4,\ b=\pm 2\)    		The distance from the center to the vertices is 4.
    The distance from the center to the endpoints of
    the minor axis is 2.
    Sketch the ellipse.

  20. ⓐ Write the equation \(6{x}^{2}+4{y}^{2}+12x-32y+34=0\) in standard form and ⓑ graph.

    답을 드러내세요

    ⓐ \(\frac{{(x+1)}^{2}}{6}+\frac{{(y-4)}^{2}}{9}=1\)
    ⓑ

  21. ⓐ Write the equation \(4{x}^{2}+{y}^{2}-16x-6y+9=0\) in standard form and ⓑ graph.

    답을 드러내세요

    ⓐ \(\frac{{(x-2)}^{2}}{4}+\frac{{(y-3)}^{2}}{16}=1\)
    ⓑ

  22. Pluto (a dwarf planet) moves in an elliptical orbit around the Sun. The closest Pluto gets to the Sun is approximately 30 astronomical units (AU) and the furthest is approximately 50 AU. The Sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of Pluto.

    답을 드러내세요

    We recognize this as an ellipse that is centered at the origin.\(\ \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\)
    Since the major axis is horizontal and the distance from the center to the vertex is 40, we know \(a=40\) and so \({a}^{2}=1600\).\(\ \frac{{x}^{2}}{1600}+\frac{{y}^{2}}{{b}^{2}}=1\)
    The minor axis is vertical but the end points aren’t given. To find \(b\) we will use the location of the Sun. Since the Sun is a focus of the ellipse at the point \((10,0)\), we know \(c=10\). Use this to solve for \({b}^{2}\).\(\ \begin{array}{l}{b}^{2}={a}^{2}-{c}^{2} \\ {b}^{2}={40}^{2}-{10}^{2} \\ {b}^{2}=1600-100 \\ {b}^{2}=1500\end{array}\)
    Substitute \({a}^{2}\) and \({b}^{2}\) into the standard form of the ellipse.\(\frac{{x}^{2}}{1600}+\frac{{y}^{2}}{1500}=1\)

  23. A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 20 AU and the furthest is approximately 30 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

    답을 드러내세요

    \(\frac{{x}^{2}}{625}+\frac{{y}^{2}}{600}=1\)

  24. A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 20 AU and the furthest is approximately 50 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

    답을 드러내세요

    \(\frac{{x}^{2}}{1225}+\frac{{y}^{2}}{1000}=1\)

  25. \(\frac{{x}^{2}}{4}+\frac{{y}^{2}}{25}=1\)

  26. \(\frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1\)

  27. \(\frac{{x}^{2}}{25}+\frac{{y}^{2}}{36}=1\)

  28. \(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{36}=1\)

  29. \(\frac{{x}^{2}}{36}+\frac{{y}^{2}}{16}=1\)

  30. \(\frac{{x}^{2}}{25}+\frac{{y}^{2}}{9}=1\)

  31. \({x}^{2}+\frac{{y}^{2}}{4}=1\)

  32. \(\frac{{x}^{2}}{9}+{y}^{2}=1\)

  33. \(4{x}^{2}+25{y}^{2}=100\)

  34. \(16{x}^{2}+9{y}^{2}=144\)

  35. \(16{x}^{2}+36{y}^{2}=576\)

  36. \(9{x}^{2}+25{y}^{2}=225\)

  37. \(\frac{{(x+1)}^{2}}{4}+\frac{{(y+6)}^{2}}{25}=1\)

  38. \(\frac{{(x-3)}^{2}}{25}+\frac{{(y+2)}^{2}}{9}=1\)

  39. \(\frac{{(x+4)}^{2}}{4}+\frac{{(y-2)}^{2}}{9}=1\)

  40. \(\frac{{(x-4)}^{2}}{9}+\frac{{(y-1)}^{2}}{16}=1\)

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어떻게: Ellipses

  1. Graph an ellipse with center at the origin
  2. Find the equation of an ellipse with center at the origin
  3. Graph an ellipse with center not at the origin
  4. Solve application with ellipses
  5. Write the equation in standard form.
  6. Determine whether the major axis is horizontal or vertical.
  7. Find the endpoints of the major axis.
  8. Find the endpoints of the minor axis

사람들이 묻는 질문

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

이 페이지의 일부는 다음에서 변경되었습니다. OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). 여기서 압축하고 다시 설명; 오류는 우리의.

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