Question

Tangent drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at point ${}^{\prime }{P}^{\prime }$ meets the coordinate axes at point $A$ and $B$ respectively. Locus of mid-night of segment $AB$ is

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

Answer 1

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

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

let the point P $\left(a\cos \theta ,b\sin \theta \right)$

$\begin{array}{c}\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1\\ A=\left(\frac{a}{\cos \theta },0\right)\&B\left(0,\frac{b}{\sin \theta }\right)\\ letthepoint\left(h,k\right)=\left(\frac{a}{2\cos \theta },\frac{b}{2\sin \theta }\right)\\ comparingthen\\ h=\frac{a}{2\cos \theta },k=\frac{b}{2\sin \theta }\\ \cos \theta =\frac{a}{2h}\&k=\frac{b}{2k}\\ now,\\ \frac{a^2}{4x^2}++\frac{b^2}{h{y}^2}=1\\ locus=\frac{a^2}{x^2}+\frac{b^2}{y^2}=4\end{array}$