Question

The locus of point of intersection of pair of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$ if the sum of ordinates of their point of contact is half the length of minor axis, is

• A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{2x}{a}=0$
• B. $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{2y}{b}=0$
• C. $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2x}{a}=0$
• D. $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2y}{b}=0$

Answer 1

The correct option is D $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2y}{b}=0$

Given, $S:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$


Let $P(h,k)$ be the point of intersection of tangents at point $A\left(\alpha \right)$ and $B\left(\beta \right).$

$\therefore P(h,k)\equiv (\frac{a\cos \frac{\alpha +\beta }{2}}{\cos \frac{\alpha -\beta }{2}},\frac{b\sin \frac{\alpha +\beta }{2}}{\cos \frac{\alpha -\beta }{2}})$

$\Rightarrow \cos \left(\frac{\alpha +\beta }{2}\right)=\frac{h}{a}\cos \left(\frac{\alpha -\beta }{2}\right)\dots \left(1\right)$
and $\sin \left(\frac{\alpha +\beta }{2}\right)=\frac{k}{b}\cos \left(\frac{\alpha -\beta }{2}\right)\dots \left(2\right)$

$\because {\cos }^2\left(\frac{\alpha +\beta }{2}\right)+{\sin }^2\left(\frac{\alpha +\beta }{2}\right)=1$

$\Rightarrow \frac{h^2}{a^2}{\cos }^2\left(\frac{\alpha -\beta }{2}\right)+\frac{k^2}{b^2}{\cos }^2\left(\frac{\alpha -\beta }{2}\right)=1$

$\Rightarrow \frac{h^2}{a^2}+\frac{k^2}{b^2}={\sec }^2\left(\frac{\alpha -\beta }{2}\right)$

Now, $b\sin \alpha +b\sin \beta =\frac{1}{2}\left(2b\right)$ (given)
$\Rightarrow \sin \alpha +\sin \beta =1$
$\Rightarrow 2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=1\dots \left(3\right)$

From $\left(2\right)$ and $\left(3\right),$ we get
$\Rightarrow {\sec }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{2k}{b}$
$\Rightarrow \frac{h^2}{a^2}+\frac{k^2}{b^2}=\frac{2k}{b}$
Hence, required locus is
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2y}{b}=0$