Question

If tangent 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 coordinate axes lie on the curve:

• 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

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

Equation of general tangent on ellipse
$\frac{x}{asec\theta }+\frac{y}{bcosec\theta }=1$

$a=\sqrt{2},b=1$
$\Rightarrow \frac{x}{\sqrt{2}sec\theta }+\frac{y}{cosec\theta }=1$
Lt the midpoint be (h,k)
$h=\frac{\sqrt{2}sec\theta }{2}\Rightarrow cos\theta =\frac{1}{\sqrt{2h}}$
and $k=\frac{cosec\theta }{2}\Rightarrow sin\theta =\frac{1}{2k}$.
$\because si{n}^2\theta +co{s}^2\theta =1$
$\Rightarrow \frac{1}{2h^2}+\frac{1}{4k^2}=1$
$\therefore \frac{1}{2x^2}+\frac{1}{4y^2}$= $1$