Question

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

• 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 $\frac{1}{2x^2}+\frac{1}{4y^2}=1$

Any tangent to the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$ is $\frac{x\cos \theta }{\sqrt{2}}+y\sin \theta =1$

The coordinates of $A$ and $B$ are

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

$\Rightarrow 2h=\sqrt{2}\sec \theta$ and $2k=\csc \theta$,

where $(h,k)$ is the mid point of $AB$

$\Rightarrow {\left(\frac{1}{\sqrt{2}h}\right)}^2+{\left(\frac{1}{2k}\right)}^2=1$

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

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