Question

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 axis 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

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

Let the point P $(\sqrt{2cos\theta },sin\theta )$ on $\frac{x^2}{2}+\frac{y^2}{1}=1$.

Equation of tangent is,
$\frac{x\sqrt{2}}{2}cos\theta +ysin\theta =1$
whose intercept on coordinate axes are
$OA=\sqrt{2}sec\theta$ and $OB=cosec\theta$
$\therefore$ Mid-point of its intercept between axes
$(\frac{\sqrt{2}}{2}sec\theta ,\frac{1}{2}cosec\theta )=(h,k)$
$\Rightarrow cos\theta =\frac{1}{\sqrt{2h}}andsin\theta =\frac{1}{2k}$
Thus, focus of mid-point M is
$(co{s}^2\theta +si{n}^2\theta )=\frac{1}{2h^2}+\frac{1}{4k^2}$
$\Rightarrow \frac{1}{2x^2}+\frac{1}{4y^2}=1$, is required locus.