Question

The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse $x^2+2y^2=2$ , between the coordinates axes, is

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

Answer 1

Answer: (C)

$\frac{1}{2x^2}+\frac{1}{4y^2}=1$

Let the point of contact be $P=(\sqrt{2}\cos \theta ,\sin \theta )$

Equation of tangent AB $=\frac{x}{\sqrt{2}}\cos \theta +y\sin \theta =1\dots \left(1\right)$

Taking $x=0$ We get $y=cosec\theta$

And taking $y=0$ We get, $x=\sqrt{2}\sec \theta$

$\Rightarrow A=(\sqrt{2}\sec \theta ,0),B=(0,\csc \theta )$

Let the middle point M of AB be $(h,k)$

$\Rightarrow h=\frac{\sqrt{2}sec\theta }{2}=\frac{\sec \theta }{\sqrt{2}},$

$k=\frac{\csc \theta }{2}$

$\Rightarrow \cos \theta =\frac{1}{h\sqrt{2}},\sin \theta =\frac{1}{2k}\dots \left(2\right)$

Substituting $\left(2\right)$ in $\left(1\right)$

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

Thus the locus of the point $M(h,k)$ is

$\Rightarrow \frac{1}{2x^2}+\frac{1}{4y^2}=1$