Question

If tangents are drawn to the ellipse $x^{2}+2 y^{2}=2$, 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. $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$
• D. $\frac{1}{4 x^{2}}+\frac{1}{2 y^{2}}=1$

Answer 1

Answer: (C)

Given ellipse,
$x^{2}+2 y^{2}=2$
$\Rightarrow \frac{x^{2}}{2}+\frac{y^{2}}{1}=1$
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 $A B$.
$\therefore h=\frac{\sqrt{2}}{2 \cos \theta}$
and $k=\frac{1}{2 \sin \theta}$
$\Rightarrow \cos \theta=\frac{1}{\sqrt{2} h}$
and $\sin \theta=\frac{1}{2 k}$
Squaring and adding, we get
$\frac{1}{2 h^{2}}+\frac{1}{4 k^{2}}=1$
$\therefore $ Locus is, $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$