Question:
Let a line $\mathrm{L}: 2 \mathrm{x}+\mathrm{y}=\mathrm{k}, \mathrm{k}>0$ be a tangent to the hyperbola $x^{2}-y^{2}=3$. If $L$ is also a tangent to the parabola $\mathrm{y}^{2}=\alpha \mathrm{x}$, then $\alpha$ is equal to :
Correct Option: , 4
Solution:
Tangent to hyperbola of
Slope $m=-2$ (given)
$y=-2 x \pm \sqrt{3(3)}$
$\left(y=m x \pm \sqrt{a^{2} m^{2}-b^{2}}\right)$
$\Rightarrow y+2 x=\pm 3 \Rightarrow 2 x+y=3 \quad(k>0)$
For parabola $\mathrm{y}^{2}-\alpha \mathrm{x}$
$\mathrm{y}=\mathrm{mx}+\frac{\alpha}{4 \mathrm{~m}}$
$\Rightarrow \mathrm{y}=-2 \mathrm{x}+\frac{\alpha}{-8}$
$\Rightarrow \frac{\alpha}{-8}=3$
$\Rightarrow \alpha=-24$