Rotation of Axes

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Find a new representation of the given equation after rotating through the given angle.

   \(3{x}^{2}+xy+3{y}^{2}-5=0,\theta =45º\)

   જવાબ બતાવો

   Find x and y using the rotation of axes formulas, substitute θ=45º.	\(\left\{\begin{array}{l}x=x'\text{cos}\theta -y'\text{sin}\theta \\ y=x'\text{sin}\theta +y'\text{cos}\theta \end{array}\right\)
   	\(x=x'\left(\frac{1}{\sqrt{2}}\right)-y'\left(\frac{1}{\sqrt{2}}\right)\) \(x=\frac{x'-y'}{\sqrt{2}}\) \(y=x'\left(\frac{1}{\sqrt{2}}\right)-y'\left(\frac{1}{\sqrt{2}}\right)\) \(y=\frac{x'-y'}{\sqrt{2}}\)
   \(x=x'\left(\frac{1}{\sqrt{2}}\right)-y'\left(\frac{1}{\sqrt{2}}\right)\) \(x=\frac{x'-y'}{\sqrt{2}}\)	\(y=x'\left(\frac{1}{\sqrt{2}}\right)-y'\left(\frac{1}{\sqrt{2}}\right)\) \(y=\frac{x'-y'}{\sqrt{2}}\)
   Substitute the expressions for x and y into the given equation and simplify.	\(3{x}^{2}+xy+3{y}^{2}-5=0\)
   	\(3{\left(\frac{x'-y'}{\sqrt{2}}\right)}^{2}+\left(\frac{x'-y'}{\sqrt{2}}\right)\left(\frac{x'-y'}{\sqrt{2}}\right)+3{\left(\frac{x'-y'}{\sqrt{2}}\right)}^{2}-5=0\)
   Foil each term.	\(3(\frac{{x'}^{2}-2x'y'+{y'}^{2}}{2})2+\frac{{x'}^{2}-{y'}^{2}}{2}+3(\frac{{x'}^{2}-2x'y'+{y'}^{2}}{2})-5=0\)
   Multiply by 2 to get rid of the fraction.	\(3({x'}^{2}-2x'y'+{y'}^{2})2+{x'}^{2}-{y'}^{2}+3({x'}^{2}-2x'y'+{y'}^{2})-10=0\)
   Combine like terms.	\(3{x'}^{2}-6x'y'+3{y'}^{2}+{x'}^{2}-{y'}^{2}+3{x'}^{2}+6x'y'+3{y'}^{2}-10=0\ 7{x'}^{2}+{5y'}^{2}-10=0\ 7{x'}^{2}+{5y'}^{2}=10\)
   Write the equations with  x′ and y′ in standard form.	Set equal to 1. \(\frac{7{x'}^{2}}{10}+\frac{{5y'}^{2}}{10}=1\ \frac{{x'}^{2}}{\frac{10}{7}}+\frac{{y'}^{2}}{2}=1\\)

2. Find a new representation of the given equation after rotating through the given angle. Use the steps outlined to assist you in your work.

   \(4{x}^{2}-xy+4{y}^{2}-2=0,\theta =45º\)

   Find x and y using the rotation of axes formulas, substitute θ=45º.

   Substitute the expressions for x and y into the given equation and simplify.

   Write the equations with  x′ and y′ in standard form.

3. 1. ⓐ \(x=-{y}^{2}-2y+3\)
   2. ⓑ \(9{y}^{2}-{x}^{2}+18y-4x-4=0\)
   3. ⓒ \(9{x}^{2}+25{y}^{2}=225\)
   4. ⓓ \({x}^{2}+{y}^{2}-4x+10y-7=0\)

   જવાબ બતાવો

   • ⓐ \(x=-{y}^{2}-2y+3\)
     Parabola: only one variable is squared.
   • ⓑ \(9{y}^{2}-{x}^{2}+18y-4x-4=0\)
     Hyperbola: \({x}^{2}\) and \({y}^{2}\) have different signs and different coefficients.
   • ⓒ \(9{x}^{2}+25{y}^{2}=225\)
     Ellipse: \({x}^{2}\) and \({y}^{2}\) have the same signs and different coefficients.
   • ⓓ \({x}^{2}+{y}^{2}-4x+10y-7=0\)
     Circle: \({x}^{2}\) and \({y}^{2}\) have the same signs and the same signs coefficients.

4. \(x=-2{y}^{2}-12y-16\)

5. \({x}^{2}+{y}^{2}=9\)

6. \(16{x}^{2}-4{y}^{2}+64x-24y-36=0\)

7. \(16{x}^{2}+36{y}^{2}=576\)

8. Identify the graph of each of the following nondegenerate conic sections.

   1. ⓐ\(4{x}^{2}-9{y}^{2}+36x+36y-125=0\)
   2. ⓑ \(9{y}^{2}+16x+36y-10=0\)
   3. ⓒ \(3{x}^{2}+3{y}^{2}-2x-6y-4=0\)
   4. ⓓ \(-25{x}^{2}-4{y}^{2}+100x+16y+20=0\)
   જવાબ બતાવો

   • ⓐ Rewriting the general form, we have

     \(A=4\) and \(C=-9,\) so we observe that \(A\) and \(C\) have opposite signs. The graph of this equation is a hyperbola.

   • ⓑ Rewriting the general form, we have

     \(A=0\) and \(C=9.\) We can determine that the equation is a parabola, since \(A\) is zero.

   • ⓒ Rewriting the general form, we have

     \(A=3\) and \(C=3.\) Because \(A=C,\) the graph of this equation is a circle.

   • ⓓ Rewriting the general form, we have

     \(A=-25\) and \(C=-4.\) Because \(AC>0\) and \(A\ne C,\) the graph of this equation is an ellipse.

9. Identify the graph of each of the following nondegenerate conic sections.

   1. ⓐ \(16{y}^{2}-{x}^{2}+x-4y-9=0\)
   2. ⓑ \(16{x}^{2}+4{y}^{2}+16x+49y-81=0\)
   જવાબ બતાવો

   1. ⓐ hyperbola
   2. ⓑ ellipse

10. Find a new representation of the equation \(2{x}^{2}-xy+2{y}^{2}-30=0\) after rotating through an angle of \(\theta =45^{\circ}.\)

    જવાબ બતાવો

    Find \(x\) and \(y,\) where \(x={x}^{'}\cos \ \theta -{y}^{'}\sin \ \theta\) and \(y={x}^{'}\sin \ \theta +{y}^{'}\cos \ \theta .\)

    Because \(\theta =45^{\circ},\)

    \[\begin{array}{l} \\ x={x}^{'}\cos (45^{\circ})-{y}^{'}\sin (45^{\circ}) \\ x={x}^{'}(\frac{1}{\sqrt{2}})-{y}^{'}(\frac{1}{\sqrt{2}}) \\ x=\frac{{x}^{'}-{y}^{'}}{\sqrt{2}}\end{array}\]

    and

    \[\begin{array}{l} \\ \begin{array}{l}y={x}^{'}\sin (45^{\circ})+{y}^{'}\cos (45^{\circ}) \\ y={x}^{'}(\frac{1}{\sqrt{2}})+{y}^{'}(\frac{1}{\sqrt{2}}) \\ y=\frac{{x}^{'}+{y}^{'}}{\sqrt{2}}\end{array}\end{array}\]

    Substitute \(x={x}^{'}\cos \theta -{y}^{'}\sin \theta\) and \(y={x}^{'}\sin \ \theta +{y}^{'}\cos \ \theta\) into \(2{x}^{2}-xy+2{y}^{2}-30=0.\)

    \[2{(\frac{{x}^{'}-{y}^{'}}{\sqrt{2}})}^{2}-(\frac{{x}^{'}-{y}^{'}}{\sqrt{2}})(\frac{{x}^{'}+{y}^{'}}{\sqrt{2}})+2{(\frac{{x}^{'}+{y}^{'}}{\sqrt{2}})}^{2}-30=0\]

    Simplify.

    \[\begin{array}{lllll}2\frac{({x}^{'}-{y}^{'})({x}^{'}-{y}^{'})}{2}-\frac{({x}^{'}-{y}^{'})({x}^{'}+{y}^{'})}{2}+2\frac{({x}^{'}+{y}^{'})({x}^{'}+{y}^{'})}{2}-30=0 & \begin{array}{llll} & & & \end{array}\text{FOIL method} \\ \ {x}^{'}{}^{2}{-2{x}^{'}y}^{'}+{y}^{'}{}^{2}-\frac{({x}^{'}{}^{2}-{y}^{'}{}^{2})}{2}+{x}^{'}{}^{2}+2{x}^{'}{y}^{'}+{y}^{'}{}^{2}-30=0 & \begin{array}{llll} & & & \end{array}\text{Combine like terms}. \\ \ 2{x}^{'}{}^{2}+2{y}^{'}{}^{2}-\frac{({x}^{'}{}^{2}-{y}^{'}{}^{2})}{2}=30 & \begin{array}{llll} & & & \end{array}\text{Combine like terms}. \\ \ 2(2{x}^{'}{}^{2}+2{y}^{'}{}^{2}-\frac{({x}^{'}{}^{2}-{y}^{'}{}^{2})}{2})=2(30) & \begin{array}{llll} & & & \end{array}\text{Multiply both sides by 2}. \\ \ 4{x}^{'}{}^{2}+4{y}^{'}{}^{2}-({x}^{'}{}^{2}-{y}^{'}{}^{2})=60 & \begin{array}{llll} & & & \end{array}\text{Simplify}. \\ \ 4{x}^{'}{}^{2}+4{y}^{'}{}^{2}-{x}^{'}{}^{2}+{y}^{'}{}^{2}=60 & \begin{array}{llll} & & & \end{array}\text{Distribute}. \\ \ \frac{3{x}^{'}{}^{2}}{60}+\frac{5{y}^{'}{}^{2}}{60}=\frac{60}{60} & \begin{array}{llll} & & & \end{array}\text{Set equal to 1}.\end{array}\]

    Write the equations with \({x}^{'}\) and \({y}^{'}\) in the standard form.

    \[\frac{{{x}^{'}}^{2}}{20}+\frac{{{y}^{'}}^{2}}{12}=1\]

    This equation is an ellipse. shows the graph.

11. Rewrite the equation \(8{x}^{2}-12xy+17{y}^{2}=20\) in the \({x}^{'}{y}^{'}\) system without an \({x}^{'}{y}^{'}\) term.

    જવાબ બતાવો

    First, we find \(\text{cot}(2\theta ).\) See .

    \[\begin{array}{l}8{x}^{2}-12xy+17{y}^{2}=20⇒A=8,\ B=-12\ \text{and}\ C=17 \\ \ \ \text{cot}(2\theta )=\frac{A-C}{B}=\frac{8-17}{-12} \\ \ \ \text{cot}(2\theta )=\frac{-9}{-12}=\frac{3}{4}\end{array}\]\[\text{cot}(2\theta )=\frac{3}{4}=\frac{\text{adjacent}}{\text{opposite}}\]

    So the hypotenuse is

    \[\begin{array}{l}{3}^{2}+{4}^{2}={h}^{2} \\ 9+16={h}^{2} \\ 25={h}^{2} \\ h=5\ \ \end{array}\]

    Next, we find \(\sin \ \theta\) and \(\cos \ \theta .\)

    \[\begin{array}{l}\begin{array}{l} \\ \\ \sin \ \theta =\sqrt{\frac{1-\cos (2\theta )}{2}}=\sqrt{\frac{1-\frac{3}{5}}{2}}=\sqrt{\frac{\frac{5}{5}-\frac{3}{5}}{2}}=\sqrt{\frac{5-3}{5}⋅\frac{1}{2}}=\sqrt{\frac{2}{10}}=\sqrt{\frac{1}{5}}\end{array} \\ \sin \ \theta =\frac{1}{\sqrt{5}} \\ \cos \ \theta =\sqrt{\frac{1+\cos (2\theta )}{2}}=\sqrt{\frac{1+\frac{3}{5}}{2}}=\sqrt{\frac{\frac{5}{5}+\frac{3}{5}}{2}}=\sqrt{\frac{5+3}{5}⋅\frac{1}{2}}=\sqrt{\frac{8}{10}}=\sqrt{\frac{4}{5}} \\ \cos \ \theta =\frac{2}{\sqrt{5}}\end{array}\]

    Substitute the values of \(\sin \ \theta\) and \(\cos \ \theta\) into \(x={x}^{'}\cos \ \theta -{y}^{'}\sin \ \theta\) and \(y={x}^{'}\sin \ \theta +{y}^{'}\cos \ \theta .\)

    \[\begin{array}{l} \\ \begin{array}{l}x={x}^{'}\cos \ \theta -{y}^{'}\sin \ \theta \\ x={x}^{'}(\frac{2}{\sqrt{5}})-{y}^{'}(\frac{1}{\sqrt{5}}) \\ x=\frac{2{x}^{'}-{y}^{'}}{\sqrt{5}}\end{array}\end{array}\]

    and

    \[\begin{array}{l}\begin{array}{l} \\ y={x}^{'}\sin \ \theta +{y}^{'}\cos \ \theta \end{array} \\ y={x}^{'}(\frac{1}{\sqrt{5}})+{y}^{'}(\frac{2}{\sqrt{5}}) \\ y=\frac{{x}^{'}+2{y}^{'}}{\sqrt{5}}\end{array}\]

    Substitute the expressions for \(x\) and \(y\) into in the given equation, and then simplify.

    \[\begin{array}{l}\ 8{(\frac{2{x}^{'}-{y}^{'}}{\sqrt{5}})}^{2}-12(\frac{2{x}^{'}-{y}^{'}}{\sqrt{5}})(\frac{{x}^{'}+2{y}^{'}}{\sqrt{5}})+17{(\frac{{x}^{'}+2{y}^{'}}{\sqrt{5}})}^{2}=20\ \\ \ 8(\frac{(2{x}^{'}-{y}^{'})(2{x}^{'}-{y}^{'})}{5})-12(\frac{(2{x}^{'}-{y}^{'})({x}^{'}+2{y}^{'})}{5})+17(\frac{({x}^{'}+2{y}^{'})({x}^{'}+2{y}^{'})}{5})=20\ \\ \ 8(4{x}^{'}{}^{2}-4{x}^{'}{y}^{'}+{y}^{'}{}^{2})-12(2{x}^{'}{}^{2}+3{x}^{'}{y}^{'}-2{y}^{'}{}^{2})+17({x}^{'}{}^{2}+4{x}^{'}{y}^{'}+4{y}^{'}{}^{2})=100 \\ 32{x}^{'}{}^{2}-32{x}^{'}{y}^{'}+8{y}^{'}{}^{2}-24{x}^{'}{}^{2}-36{x}^{'}{y}^{'}+24{y}^{'}{}^{2}+17{x}^{'}{}^{2}+68{x}^{'}{y}^{'}+68{y}^{'}{}^{2}=100 \\ \ 25{x}^{'}{}^{2}+100{y}^{'}{}^{2}=100\ \\ \ \frac{25}{100}{x}^{'}{}^{2}+\frac{100}{100}{y}^{'}{}^{2}=\frac{100}{100}\end{array}\]

    Write the equations with \({x}^{'}\) and \({y}^{'}\) in the standard form with respect to the new coordinate system.

    \[\frac{{{x}^{'}}^{2}}{4}+\frac{{{y}^{'}}^{2}}{1}=1\]

    shows the graph of the ellipse.

12. Rewrite the \(13{x}^{2}-6\sqrt{3}xy+7{y}^{2}=16\) in the \({x}^{'}{y}^{'}\) system without the \({x}^{'}{y}^{'}\) term.

    જવાબ બતાવો

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

13. Graph the following equation relative to the \({x}^{'}{y}^{'}\) system:

    \[{x}^{2}+12xy-4{y}^{2}=30\]
    જવાબ બતાવો

    First, we find \(\text{cot}(2\theta ).\)

    \[{x}^{2}+12xy-4{y}^{2}=20⇒A=1,\ B=12,\text{and }C=-4\]\[\begin{array}{l}\text{cot}(2\theta )=\frac{A-C}{B} \\ \text{cot}(2\theta )=\frac{1-(-4)}{12} \\ \text{cot}(2\theta )=\frac{5}{12}\end{array}\]

    Because \(\text{cot}(2\theta )=\frac{5}{12},\) we can draw a reference triangle as in .

    \[\text{cot}(2\theta )=\frac{5}{12}=\frac{\text{adjacent}}{\text{opposite}}\]

    Thus, the hypotenuse is

    \[\begin{array}{l}{5}^{2}+{12}^{2}={h}^{2} \\ 25+144={h}^{2} \\ 169={h}^{2} \\ h=13\end{array}\]

    Next, we find \(\sin \ \theta\) and \(\cos \ \theta .\) We will use half-angle identities.

    \[\begin{array}{l}\begin{array}{l} \\ \\ \sin \ \theta =\sqrt{\frac{1-\cos (2\theta )}{2}}=\sqrt{\frac{1-\frac{5}{13}}{2}}=\sqrt{\frac{\frac{13}{13}-\frac{5}{13}}{2}}=\sqrt{\frac{8}{13}⋅\frac{1}{2}}=\frac{2}{\sqrt{13}}\end{array} \\ \cos \ \theta =\sqrt{\frac{1+\cos (2\theta )}{2}}=\sqrt{\frac{1+\frac{5}{13}}{2}}=\sqrt{\frac{\frac{13}{13}+\frac{5}{13}}{2}}=\sqrt{\frac{18}{13}⋅\frac{1}{2}}=\frac{3}{\sqrt{13}}\end{array}\]

    Now we find \(x\) and \(y\text{.\,}\)

    \[\begin{array}{l} \\ x={x}^{'}\cos \ \theta -{y}^{'}\sin \ \theta \\ x={x}^{'}(\frac{3}{\sqrt{13}})-{y}^{'}(\frac{2}{\sqrt{13}}) \\ x=\frac{3{x}^{'}-2{y}^{'}}{\sqrt{13}}\end{array}\]

    and

    \[\begin{array}{l} \\ y={x}^{'}\sin \ \theta +{y}^{'}\cos \ \theta \\ y={x}^{'}(\frac{2}{\sqrt{13}})+{y}^{'}(\frac{3}{\sqrt{13}}) \\ y=\frac{2{x}^{'}+3{y}^{'}}{\sqrt{13}}\end{array}\]

    Now we substitute \(x=\frac{3{x}^{'}-2{y}^{'}}{\sqrt{13}}\) and \(y=\frac{2{x}^{'}+3{y}^{'}}{\sqrt{13}}\) into \({x}^{2}+12xy-4{y}^{2}=30.\)

    \[\begin{array}{llll}\ {(\frac{3{x}^{'}-2{y}^{'}}{\sqrt{13}})}^{2}+12(\frac{3{x}^{'}-2{y}^{'}}{\sqrt{13}})(\frac{2{x}^{'}+3{y}^{'}}{\sqrt{13}})-4{(\frac{2{x}^{'}+3{y}^{'}}{\sqrt{13}})}^{2}=30 & & & \\ \ \ (\frac{1}{13})[{(3{x}^{'}-2{y}^{'})}^{2}+12(3{x}^{'}-2{y}^{'})(2{x}^{'}+3{y}^{'})-4{(2{x}^{'}+3{y}^{'})}^{2}]=30 & & & \text{Factor}. \\ (\frac{1}{13})[9{x}^{'}{}^{2}-12{x}^{'}{y}^{'}+4{y}^{'}{}^{2}+12(6{x}^{'}{}^{2}+5{x}^{'}{y}^{'}-6{y}^{'}{}^{2})-4(4{x}^{'}{}^{2}+12{x}^{'}{y}^{'}+9{y}^{'}{}^{2})]=30 & & & \text{Multiply}. \\ \ (\frac{1}{13})[9{x}^{'}{}^{2}-12{x}^{'}{y}^{'}+4{y}^{'}{}^{2}+72{x}^{'}{}^{2}+60{x}^{'}{y}^{'}-72{y}^{'}{}^{2}-16{x}^{'}{}^{2}-48{x}^{'}{y}^{'}-36{y}^{'}{}^{2}]=30 & & & \text{Distribute}. \\ \ \ (\frac{1}{13})[65{x}^{'}{}^{2}-104{y}^{'}{}^{2}]=30 & & & \text{Combine like terms}. \\ \ 65{x}^{'}{}^{2}-104{y}^{'}{}^{2}=390 & & & \text{Multiply}.\ \\ \ \frac{{x}^{'}{}^{2}}{6}-\frac{4{y}^{'}{}^{2}}{15}=1 & & & \text{Divide by 390}.\end{array}\]

    shows the graph of the hyperbola \(\frac{{{x}^{'}}^{2}}{6}-\frac{4{{y}^{'}}^{2}}{15}=1.\ \\)

14. Identify the conic for each of the following without rotating axes.

    1. ⓐ \(5{x}^{2}+2\sqrt{3}xy+2{y}^{2}-5=0\)
    2. ⓑ \(5{x}^{2}+2\sqrt{3}xy+12{y}^{2}-5=0\)
    જવાબ બતાવો

    • ⓐ Let’s begin by determining \(A,B,\) and \(C.\) \[\underset{A}{\underset{︸}{5}}{x}^{2}+\underset{B}{\underset{︸}{2\sqrt{3}}}xy+\underset{C}{\underset{︸}{2}}{y}^{2}-5=0\]

      Now, we find the discriminant.

      \[\begin{array}{l}{B}^{2}-4AC={(2\sqrt{3})}^{2}-4(5)(2) \\ \ =4(3)-40 \\ \ =12-40 \\ \ =-28<0\end{array}\]

      Therefore, \(5{x}^{2}+2\sqrt{3}xy+2{y}^{2}-5=0\) represents an ellipse.

    • ⓑ Again, let’s begin by determining \(A,B,\) and \(C.\) \[\underset{A}{\underset{︸}{5}}{x}^{2}+\underset{B}{\underset{︸}{2\sqrt{3}}}xy+\underset{C}{\underset{︸}{12}}{y}^{2}-5=0\]

      Now, we find the discriminant.

      \[\begin{array}{l}{B}^{2}-4AC={(2\sqrt{3})}^{2}-4(5)(12) \\ \ =4(3)-240 \\ \ =12-240 \\ \ =-228<0\end{array}\]

      Therefore, \(5{x}^{2}+2\sqrt{3}xy+12{y}^{2}-5=0\) represents an ellipse.

15. Identify the conic for each of the following without rotating axes.

    1. ⓐ \({x}^{2}-9xy+3{y}^{2}-12=0\)
    2. ⓑ \(10{x}^{2}-9xy+4{y}^{2}-4=0\)
    જવાબ બતાવો

    1. ⓐ hyperbola
    2. ⓑ ellipse

16. What effect does the \(xy\) term have on the graph of a conic section?

    જવાબ બતાવો

    The \(xy\) term causes a rotation of the graph to occur.

17. If the equation of a conic section is written in the form \(A{x}^{2}+B{y}^{2}+Cx+Dy+E=0\) and \(AB=0,\) what can we conclude?

18. If the equation of a conic section is written in the form \(A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0,\) and \({B}^{2}-4AC>0,\) what can we conclude?

    જવાબ બતાવો

    The conic section is a hyperbola.

19. Given the equation \(a{x}^{2}+4x+3{y}^{2}-12=0,\) what can we conclude if \(a>0?\)

20. For the equation \(A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0,\) the value of \(\theta\) that satisfies \(\text{cot}(2\theta )=\frac{A-C}{B}\) gives us what information?

    જવાબ બતાવો

    It gives the angle of rotation of the axes in order to eliminate the \(xy\) term.

21. \(9{x}^{2}+4{y}^{2}+72x+36y-500=0\)

22. \({x}^{2}-10x+4y-10=0\)

    જવાબ બતાવો

    \(AB=0,\) parabola

23. \(2{x}^{2}-2{y}^{2}+4x-6y-2=0\)

24. \(4{x}^{2}-{y}^{2}+8x-1=0\)

    જવાબ બતાવો

    \(AB=-4<0,\) hyperbola

25. \(4{y}^{2}-5x+9y+1=0\)

26. \(2{x}^{2}+3{y}^{2}-8x-12y+2=0\)

    જવાબ બતાવો

    \(AB=6>0,\) ellipse

27. \(4{x}^{2}+9xy+4{y}^{2}-36y-125=0\)

28. \(3{x}^{2}+6xy+3{y}^{2}-36y-125=0\)

    જવાબ બતાવો

    \({B}^{2}-4AC=0,\) parabola

29. \(-3{x}^{2}+3\sqrt{3}xy-4{y}^{2}+9=0\)

30. \(2{x}^{2}+4\sqrt{3}xy+6{y}^{2}-6x-3=0\)

    જવાબ બતાવો

    \({B}^{2}-4AC=0,\) parabola

31. \(-{x}^{2}+4\sqrt{2}xy+2{y}^{2}-2y+1=0\)

32. \(8{x}^{2}+4\sqrt{2}xy+4{y}^{2}-10x+1=0\)

    જવાબ બતાવો

    \({B}^{2}-4AC=-96<0,\) ellipse

33. \(3{x}^{2}+xy+3{y}^{2}-5=0,\theta =45^{\circ}\)

34. \(4{x}^{2}-xy+4{y}^{2}-2=0,\theta =45^{\circ}\)

    જવાબ બતાવો

    \(7{{x}^{'}}^{2}+9{{y}^{'}}^{2}-4=0\)

35. \(2{x}^{2}+8xy-1=0,\theta =30^{\circ}\)

36. \(-2{x}^{2}+8xy+1=0,\theta =45^{\circ}\)

    જવાબ બતાવો

    \(3{{x}^{'}}^{2}+2{x}^{'}{y}^{'}-5{{y}^{'}}^{2}+1=0\)

37. \(4{x}^{2}+\sqrt{2}xy+4{y}^{2}+y+2=0,\theta =45^{\circ}\)

38. \({x}^{2}+3\sqrt{3}xy+4{y}^{2}+y-2=0\)

    જવાબ બતાવો

    \(\theta ={60}^{∘},11{{x}^{'}}^{2}-{{y}^{'}}^{2}+\sqrt{3}{x}^{'}+{y}^{'}-4=0\)

39. \(4{x}^{2}+2\sqrt{3}xy+6{y}^{2}+y-2=0\)

40. \(9{x}^{2}-3\sqrt{3}xy+6{y}^{2}+4y-3=0\)

    જવાબ બતાવો

    \(\theta ={-30}^{∘},21{{x}^{'}}^{2}+9{{y}^{'}}^{2}+4{x}^{'}-4\sqrt{3}{y}^{'}-6=0\)

તમારા પોતાના પ્રયત્ન કરો

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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