Question

Tangents are drawn at right angles to the $ellipse\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then the locus of mid points of chords of contact is

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

Answer 1

The correct option is B $\frac{x^2+y^2}{a^2+b^2}={(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2$

Since, tangents are drawn at right angles
Therefore, locus of point of intersection of tangents is a director circle $x^2+y^2=a^2+b^2$
Let any point on director circle be $(\sqrt{a^2+b^2}\cos \theta ,\sqrt{a^2+b^2}\sin \theta )$
Then equation of chord of contact is $T=0$
$\frac{x\sqrt{a^2+b^2}\cos \theta }{a^2}+\frac{y\sqrt{a^2+b^2}\sin \theta }{b^2}=1$ ......(1)
now, let $(h,k)$ be the mid point of chord of contact
Then equation of chord is $T=S_1$
$\frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$ .....(2)
Comparing (1) and (2): $\frac{h}{\sqrt{a^2+b^2}\cos \theta }=\frac{k}{\sqrt{a^2+b^2}\sin \theta }=\frac{h^2}{a^2}+\frac{k^2}{b^2}$
eliminating $\theta$, we get
$\frac{x^2+y^2}{a^2+b^2}={(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2$
Ans: B