Question

If the chord of contact of tangents drawn from a point $P$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ subtends a right-angle at its centre, then $P$ lies on:

• A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{a^2}+\frac{1}{b^2}$
• B. $\frac{x^2}{a^4}+\frac{y^2}{b^4}={(\frac{1}{a}+\frac{1}{b})}^2$
• C. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{a^4}+\frac{1}{b^4}$
• D. $\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}$

Answer 1

The correct option is D $\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}$

Let the point P be $(h,k)$.

Equation of chord of contact is: $\frac{hx}{a^2}+\frac{ky}{b^2}=1$ -----(1)

Equation of Ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ----(2)

Homogeneous ellipse with the help of equation (1):

$\frac{x^2}{a^2}+\frac{y^2}{b^2}={(\frac{hx}{a^2}+\frac{ky}{b^2})}^2$

$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{{\left(hx\right)}^2}{a^4}+\frac{{\left(ky\right)}^2}{b^4}+2\frac{khyx}{a^2{b}^2}$

Now, equating the co-efficient of $x^2+y^{2}to0$

As, it subtends right angle at origin:

$\frac{h^2}{a^4}+\frac{k^2}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}$

The required locus is:$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}$