Question

Find the locus of the mid-points of chords of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the tangents at the extremities of which intersect at right angles.

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

Answer 1

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

Let $(x_1,y_1)$ be the point of intersection of perpendicular tangents so that $(x_1,y_1)$ lies on director circle $\therefore x_1^2+y_1^2=a^2+b^2...\left(1\right)$
Then the chord will be chord of contact of $(x_1,y_1)$

$\therefore \frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=\mathrm{1...}\left(2\right)$
If its mid-point is $(h,k)$, then its equation is $\frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}...\left(3\right)$
Compare (2) and (3) and find $x_1,y_1$ and put in (1)
$\therefore$ Locus of $(h,k)$ is ${(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2(a^2+b^{{}^2})=x^2+y^2$
Ans: D