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

The correct option is C $(x^2+y^2{)}^2=6x^2+2y^2$

Let the foot of perpendicular be $P(h,k)$

Equation of tangent with slope $m$ passing $P(h,k)$ is

$y=mx\pm \sqrt{6m^2+2},$ where $m=-\frac{h}{k}$

$\Rightarrow \sqrt{\frac{6h^2}{k^2}+2}=\frac{h^2+k^2}{k}$

$6h^2+2k^2=(h^2+k^2{)}^2$

So required locus is $6x^2+2y^2=(x^2+y^2{)}^2$