Question

If tangents are drawn to the ellipse $x^2+2y^2=2$, then the locus of the mid points of the intercepts made by those tangents between the coordinate axes is

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

Answer 1

Answer: (C)

Given ellipse,
$x^2+2y^2=2$
$\Rightarrow \frac{x^2}{2}+\frac{y^2}{1}=1$
Point $P(\sqrt{2}\cos \theta ,\sin \theta )$ lie in ellipse
Tangent at $P(\sqrt{2}\cos \theta ,\sin \theta )$ on ellipse is,
$x\cos +\sqrt{2}\sin \theta y=\sqrt{2}$
Intercept on line is, $A\left(\frac{\sqrt{2}}{\cos \theta },0\right)$
and $B\left(0,\frac{1}{\sin \theta }\right)$
Let $(h,k)$ is mid-point of the $AB$.
$\therefore h=\frac{\sqrt{2}}{2\cos \theta }$
and $k=\frac{1}{2\sin \theta }$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}h}$
and $\sin \theta =\frac{1}{2k}$
Squaring and adding, we get
$\frac{1}{2h^2}+\frac{1}{4k^2}=1$
$\therefore$ Locus is, $\frac{1}{2x^2}+\frac{1}{4y^2}=1$