Question

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

• A. $(x^2+y^2{)}^2=6x^2+2y^2$
• B. $(x^2+y^2{)}^2=6x^2-2y^2$
• C. $(x^2-y^2{)}^2=6x^2+2y^2$
• D. $(x^2-y^2{)}^2=6x^2-2y^2$

Answer 1

Answer: (A)

Equation of ellipse is $x^2+3y^2=6$ or $\frac{x^2}{6}+\frac{y^2}{2}=1$.
Equation of the tangent is $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
Let $(h,k)$ be any point on the locus.
$\therefore \frac{h}{a}\cos \theta +\frac{k}{b}\sin \theta =\mathrm{1....}$ (i)
Slope of the tangent line is $\frac{-b}{a}\cot \theta$.
Slope of perpendicular drawn from centre $(0,0)$ to $(h,k)$ is $k/h$.
Since, both the lines are perpendicular.
$\therefore \left(\frac{k}{h}\right)\times \left(-\frac{b}{a}\cot \theta \right)=-1$
$\Rightarrow \frac{\cos \theta }{ha}=\frac{\sin \theta }{kb}=\alpha$ [say]
$\Rightarrow \cos \theta =\alpha ha$
$\sin \theta =\alpha kb$
From Eq. (i), $\frac{h}{a}(\alpha ha)+\frac{k}{b}(\alpha kb)=1$
$\Rightarrow h^2\alpha +k^2\alpha =1$
$\Rightarrow \alpha =\frac{1}{h^2+k^2}$
Also,${\sin }^2\theta +{\cos }^2\theta =1$
$\Rightarrow (\alpha kb{)}^2+(\alpha ha{)}^2=1$
$\Rightarrow {\alpha }^2{k}^2{b}^2+{\alpha }^2{h}^2{a}^2=1$
$\Rightarrow \frac{k^2{b}^2}{{\left(h^2+k^2\right)}^2}+\frac{h^2{a}^2}{{\left(h^2+k^2\right)}^2}=1$
$\Rightarrow \frac{2k^2}{{\left(h^2+k^2\right)}^2}+\frac{6h^2}{{\left(h^2+k^2\right)}^2}=1$
$\left[\because a^2=6,b^2=2\right]$
$\Rightarrow 6x^2+2y^2={\left(x^2+y^2\right)}^2$
[replacing $k$ by $y$ and $h$ by $x$]