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: (C)

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

The given ellipse is $x^2+3y^2=6$

$\Rightarrow \frac{x^2}{6}+\frac{y^2}{2}=1$

Now we know that any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by

$y=mx+\sqrt{a^2{m}^2+b^2}$

So, the equation of the tangent to the given ellipse is

$y=mx+\sqrt{6m^2+2}$.......$\left(1\right)$

Now,equation of the line through $(0,0)$ and perpendicular to $\left(1\right)$ is

$y-0=\left(\frac{-1}{m}\right)x-0$

$\Rightarrow m=-\frac{x}{y}$.....$\left(2\right)$

Using $\left(2\right)$ in $\left(1\right)$, we have

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

$\Rightarrow y^2=-x^2+\sqrt{6x^2+2y^2}$

$\Rightarrow y^2+x^2=\sqrt{6x^2+2y^2}$

(or) ${(y^2+x^2)}^2=6x^2+2y^2$ is the required locus.