# Trace the following central conics:

${y^2} - 2xy + 2{x^2} + 2x - 2y = 0$

Answer

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Hint: -We know equation of general equation of second degree is $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ and we also know about $\Delta = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$ and you should know that if $\Delta = 0$ it will be straight line, but if not then check many things like center of curve eccentricity of conic and many more to trace.

Complete step-by-step answer:

We have given ${y^2} - 2xy + 2{x^2} + 2x - 2y = 0$

On comparing with general equation of second degree we get,

So a =2, b =1, h = -1, g = 1, f = -1

We know

$\Delta = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$

$

\Delta = 0 + 2 - 2 - 1 = - 1 \\

\Delta \ne 0 \\

{h^2} = 1 \\

ab = 2 \\

{h^2} < ab \\

\\

$

(you should have knowledge that if ${h^2} = ab$ it will be parabola and if ${h^2} < ab$ it will be ellipse and if ${h^2} > ab$ it will be hyperbola)

Here you can directly say this is the equation of the ellipse from the above note.

Now we have to find some more information to trace the ellipse,

If you have to find the center of the curve first you have to partially differentiate with respect to x and then with respect to y, and solve both the equations to get the center of the curve.

$

\dfrac{\partial }{{\partial y}}\left( {{y^2} - 2xy + 2{x^2} + 2x - 2y = 0} \right) \\

4x - 2y + 2 = 0 \\

2x - y + 1 = 0 \ldots \left( i \right) \\

$

$

\dfrac{\partial }{{\partial y}}\left( {{y^2} - 2xy + 2{x^2} + 2x - 2y = 0} \right) \\

2y - 2x - 2 = 0 \\

y - x - 1 = 0 \ldots \left( {ii} \right) \\

$

On solving equation $\left( i \right)$ and $\left( {ii} \right)$ we get

X=0, y =1

C (0,1)

We know

$

\tan 2\theta = \dfrac{{2h}}{{a - b}} \\

\tan 2\theta = \dfrac{{ - 2}}{{2 - 1}} = - 2 \\

\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = - 2 \\

2{\tan ^2}\theta - 2\tan \theta - 2 = 0 \\

{\tan ^2}\theta - \tan \theta - 1 = 0 \\

\tan \theta = \dfrac{{1 \pm \sqrt {1 + 4} }}{2} \\

\tan \theta = \dfrac{{1 \pm \sqrt 5 }}{2} \\

$

Is the position of the axes of the ellipse.

Note: -To solve this type of question you have knowledge of the general equation of second degree. You have to find the center and eccentricity to trace the curve and also angle to find the position of the axis.

Complete step-by-step answer:

We have given ${y^2} - 2xy + 2{x^2} + 2x - 2y = 0$

On comparing with general equation of second degree we get,

So a =2, b =1, h = -1, g = 1, f = -1

We know

$\Delta = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$

$

\Delta = 0 + 2 - 2 - 1 = - 1 \\

\Delta \ne 0 \\

{h^2} = 1 \\

ab = 2 \\

{h^2} < ab \\

\\

$

(you should have knowledge that if ${h^2} = ab$ it will be parabola and if ${h^2} < ab$ it will be ellipse and if ${h^2} > ab$ it will be hyperbola)

Here you can directly say this is the equation of the ellipse from the above note.

Now we have to find some more information to trace the ellipse,

If you have to find the center of the curve first you have to partially differentiate with respect to x and then with respect to y, and solve both the equations to get the center of the curve.

$

\dfrac{\partial }{{\partial y}}\left( {{y^2} - 2xy + 2{x^2} + 2x - 2y = 0} \right) \\

4x - 2y + 2 = 0 \\

2x - y + 1 = 0 \ldots \left( i \right) \\

$

$

\dfrac{\partial }{{\partial y}}\left( {{y^2} - 2xy + 2{x^2} + 2x - 2y = 0} \right) \\

2y - 2x - 2 = 0 \\

y - x - 1 = 0 \ldots \left( {ii} \right) \\

$

On solving equation $\left( i \right)$ and $\left( {ii} \right)$ we get

X=0, y =1

C (0,1)

We know

$

\tan 2\theta = \dfrac{{2h}}{{a - b}} \\

\tan 2\theta = \dfrac{{ - 2}}{{2 - 1}} = - 2 \\

\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = - 2 \\

2{\tan ^2}\theta - 2\tan \theta - 2 = 0 \\

{\tan ^2}\theta - \tan \theta - 1 = 0 \\

\tan \theta = \dfrac{{1 \pm \sqrt {1 + 4} }}{2} \\

\tan \theta = \dfrac{{1 \pm \sqrt 5 }}{2} \\

$

Is the position of the axes of the ellipse.

Note: -To solve this type of question you have knowledge of the general equation of second degree. You have to find the center and eccentricity to trace the curve and also angle to find the position of the axis.

Last updated date: 24th Sep 2023

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