Question

If a pair of variable straight lines $x^2+4y^2+\alpha xy=0$ (where $\alpha$ is a real parameter) cut the ellipse $x^2+4y^2=4$ at two points $A$ and $B$, then the locus of the point of intersection of tangents at $A$ and $B$ is

• A. $x^2-4y^2+8xy=0$
• B. $(2x-y)(2x+y)=0$
• C. $x^2-4y^2+4xy=0$
• D. $(x-2y)(x+2y)=0$

Answer 1

The correct option is D $(x-2y)(x+2y)=0$

Let the point of intersection of tangents at $A$ and $B$ be $P(h,k)$, then
Equation of $AB$ is $\frac{xh}{4}+\frac{yk}{1}=1$ $....\left(1\right)$
Homogenizing the ellipse using $\left(1\right)$

$\frac{x^2}{4}+\frac{y^2}{1}={(\frac{xh}{4}+\frac{yk}{1})}^2$
$\Rightarrow x^2\left(\frac{h^2-4}{16}\right)+y^2(k^2-1)+\frac{2hk}{4}xy=0$ $....\left(2\right)$
Given equation of $OA$ and $OB$ is

$x^2+4y^2+\alpha xy=\mathrm{0....}\left(3\right)$
$\because$ (2) and (3) represent same line.

Hence,
$\frac{h^2-4}{16}=\frac{k^2-1}{4}=\frac{hk}{2\alpha }$
$h^2-4=4(k^2-1)\Rightarrow h^2-4k^2=0$
$(h-2k)(h+2k)=0$

$\therefore$ Locus is $(x-2y)(x+2y)=0$