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

Answer 1

The correct option is D

$(x-2y)(x+2y)=0$

Step 1: Write equation for $AB$

Let the locus for intersection of tangents at $A,B$ be $P(h,k)$

Equation of $AB$ is given as $\frac{xh}{a}+\frac{yk}{b}=1$

Here $a=4,b=1$

$\therefore$ $AB:\frac{xh}{4}+\frac{yk}{1}=1$ $...\left(i\right)$

Step 2: Homogenise the ellipse

$\frac{x^2}{4}+\frac{y^2}{1}={\left(\frac{xh}{4}+\frac{yk}{1}\right)}^2$

$\Rightarrow$ $x^2\left(\frac{h^2-4}{16}\right)+y^2\left(k^2-1\right)+hk\frac{xy}{2}=0$ $...\left(ii\right)$

Given pair of straight lines is $x^2+4y^2+\alpha xy=0$ $...\left(iii\right)$

Step 3 : Solve for $h,k$

Equations $\left(ii\right),\left(iii\right)$ represent the same lines.

Comparing the coefficients we get

$\frac{h^2-4}{16}=\frac{k^2-1}{4}=\frac{hk}{2\alpha }$

$\Rightarrow$ $h^2-4=4(k^2-1)$

$\Rightarrow$ $h^2-4k^2=0$

$\Rightarrow$ $\left(h+2k\right)\left(h-2k\right)=0$

Replace $h,k$ with $x,y$ to obtain the locus

$\therefore$ $\left(x+2y\right)\left(x-2y\right)=0$ is the locus of point

Hence, option $\left(D\right)$ is the correct answer