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 drawn from the centre of ellipse on the tangent ${}^{\prime }{l}^{\prime }$ is $(h,k)$


Slope of $OP=\frac{k}{h}$
$\therefore$ Slope of tangent, $m=-\frac{h}{k}$
As we know, Equation of tangent is
$y=mx\pm \sqrt{a^2{m}^2+b^2}...\left(1\right)$
put value of $m$ in equation (1) and pass through $(h,k)$
Thus, $k=-\frac{h}{k}\cdot h\pm \sqrt{6{(-\frac{h}{k})}^2+2}$
$\Rightarrow k^2+h^2=\pm \sqrt{6h^2+2k^2}$
$\Rightarrow (k^2+h^2{)}^2=6h^2+2k^2$
Put (x,y) at the place of (h,k)
Hence, $(x^2+y^2{)}^2=6x^2+2y^2$