Question

If tangents are drawn to the ellipse $x^2+2y^2=2$ at all points on the ellipse other than its four vertices, then the mid points of the tangents intercepted between the coordinates axes lie on the curve :

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

Answer 1

The correct option is A $\frac{1}{2x^2}+\frac{1}{4y^2}=1$

Equation of tangent to ellipse $x^2+2y^2=2$ at $(\sqrt{2}\cos \theta ,\sin \theta )$ is $\frac{x\cos \theta }{\sqrt{2}}+\frac{y\sin \theta }{1}=1$
Therefore, it intersects co-ordinate axes at $(\frac{\sqrt{2}}{\cos \theta },0)$ and $(0,\frac{1}{\sin \theta })$
Mid point $(h,k)$ is
$h=\frac{1}{\sqrt{2}\cos \theta }\Rightarrow \cos \theta =\frac{1}{\sqrt{2}h}$
$k=\frac{1}{2\sin \theta }\Rightarrow \sin \theta =\frac{1}{2k}$

${\sin }^2\theta +{\cos }^2\theta =1$
$\Rightarrow \frac{1}{2h^2}+\frac{1}{4k^2}=1$
$\therefore$ Locus is $\frac{1}{2x^2}+\frac{1}{4y^2}=1$