The locus of the foot of perpendicular drawn | vedclass

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Question

$x^{2}+3 y^{2}=6$

$\Rightarrow \frac{x^{2}}{6}+\frac{y^{2}}{2}=1$

Now, we know that any tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}

Vedclass · Accepted Answer

${\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$

Vedclass · Answer

$x^{2}+3 y^{2}=6$

$\Rightarrow \frac{x^{2}}{6}+\frac{y^{2}}{2}=1$

Now, we know that any tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is given by:

$y=m x+\sqrt{a^{2} m^{2}+b^{2}}$

So, the equation of the tangent to the given ellipse is:-

$y=m x+\sqrt{6 m^{2}+2}$          ...........$(1)$

Now, equation of the line through $(0,0)$ and perpendicular to $(1)$ is:-

$y-0=\left(-\frac{1}{m}\right) x-0$

$\Rightarrow m=-\frac{x}{y}$          .....$(2)$

Using $( 2)$ in $(1),$ we have:

$y=\left(-\frac{x}{y}\right) x+\sqrt{6\left(\frac{x^{2}}{y^{2}}\right)+2}$

$\Rightarrow y^{2}=-x^{2}+\sqrt{6 x^{2}+2 y^{2}}$

$\Rightarrow\left(x^{2}+y^{2}\right)^{2}=6 x^{2}+2 y^{2}$

The locus of the foot of perpendicular drawn from the centre of the ellipse ${x^2} + 3{y^2} = 6$ on any tangent to it is

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