Question

If a variable tangent to the circles $x^2+y^2=1$ intersects the ellipse $x^2+{2y}^2=4$ at points P and Q then the locus of the point of intersection of tangents to the ellipse at P and Q is a conic whose

• A. eccentricity $\frac{\sqrt{3}}{4}$
• B. eccentricity is $\frac{\sqrt{5}}{2}$
• C. latus-rectum is of length 2 units
• D. foci are $(\pm 2\sqrt{5},0)$

Answer 1

The correct option is A eccentricity $\frac{\sqrt{3}}{4}$

Given,

$x^2+y^2=1$

$x^2+2y^2=4$

from the above equations, we have,

$hx+2ky=4$

$\to \left|\frac{-4}{\sqrt{h^2+4k^2}}\right|=1$

$\Rightarrow h^2+4k^2=16$

$\Rightarrow x^2+4y^2=16$

$\Rightarrow \frac{x^2}{16}+\frac{y^2}{4}=1$

eccentricity,

$e=\sqrt{1-\frac{4}{16}}=\frac{\sqrt{3}}{4}$

Length of latus rectum,

$l=\frac{2b^2}{a}=\frac{2\times 4}{4}=2$