The locus of the mid points of the intercepted portion of the tangents by the coordinate axes, which are drawn to the ellipse $x^2+2 y^2=2$ is

• Correct answer
  A

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

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

  $\frac{x^2}{2}+\frac{y^2}{4}=1$
• D

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

Solution

$$ \frac{x^2}{2}+\frac{y^2}{1}=1 \Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$ Equation of tangent at $P(a \cos \theta, b \sin \theta)$
$\begin{aligned} & \frac{x}{\sqrt{2}} \cos \theta+y \sin \theta=1, \text { For } A, y=0 \\ & x=\sqrt{2} \sec \theta, \text { For } B, x=0, y=\operatorname{cosec} \theta \\ & \therefore \quad \text { Midpoint of } A B=\left(\frac{x}{2}, \frac{y}{2}\right) \\ & M \equiv\left(\frac{\sec \theta}{\sqrt{2}}, \frac{\operatorname{cosec} \theta}{2}\right) \equiv\left(\left(\frac{x}{2}\right),\left(\frac{y}{2}\right)\right) \equiv(h, k) \\ & \because \quad \sin ^2 \theta+\cos ^2 \theta=-1 \\ & \Rightarrow \quad\left(\frac{1}{2 k}\right)^2+\left(\frac{1}{\sqrt{2} h}\right)^2=1 \Rightarrow \frac{1}{4 k^2}+\frac{1}{2 h^2}=1 \\ & \therefore \quad \frac{1}{2 x^2}+\frac{1}{4 y^2}=1\end{aligned}$

Hence, the correct answer is option: 'A'.