If a pair of variable straight lines x² + 4y² + αxy = 0 (where α is a real parameter) cut the ellipse x² + 4y² = 4 at two points A and B, then the locus of the point of intersection of tangents at A and B is

Given ellipse $x^2+4y^2=1\to \left(1\right)$

Let P(x₁ y₁) be the point of intersection of tangents at A and B
Equation of chord of contact of P is xx₁ + 4yy₁ = 4 .......(2)
Homogenizing (1) w.r.t (2)

$x^2+4y^2-1{\left(\frac{x{x}_1+4y{y}_1}{4}\right)}^2=0$
$\Rightarrow \left(4-{\mathrm{x}}_1^2\right){\mathrm{x}}^2-8{\mathrm{x}}_1{\mathrm{y}}_1\cdot \mathrm{xy}+16\left(1-{\mathrm{y}}_1^2\right){\mathrm{y}}^2=0$
But the given pair of lines is
$\begin{array}{l}{\mathrm{x}}^2+\mathrm{\alpha xy}+4{\mathrm{y}}^2=0\\ By comparision we get 4-{\mathrm{x}}_1^2=1,4\left(1-{\mathrm{y}}^2\right)=1\\ {{\mathrm{x}}_1}^2=3;{{\mathrm{y}}_1}^2=\frac{3}{4}\\ \therefore \mathrm{the} \mathrm{points} \mathrm{are} \left(\pm \sqrt{3},\pm \frac{\sqrt{3}}{2}\right)\end{array}$
$\Rightarrow$These are satisfies the equations  x -2y= 0, x+ 2y= 0