- A$\frac {1}{3}$
- B$\frac {1}{\sqrt 3}$
- C$\sqrt 3$
- ✓$3$
${\tan ^2}\alpha {\left( {x + a} \right)^2} = {y^2} - 4ax$
Given equation of Parabola
${y^2} - 4a\left\{ {a = 1} \right\}$
Point of intersection $\left( { - 2, - 1} \right)$
${\tan ^2}\alpha {\left( { - 2 + 1} \right)^2} = {\left( { - 1} \right)^2} - 4 \times 1 \times \left( { - 2} \right)$
$ \Rightarrow {\tan ^2}\alpha = 9$
$ \Rightarrow \tan \alpha = \pm 3$
$ \Rightarrow \left| {\tan \alpha } \right| = 3$
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Statement $-1$:An equation of a common tangent to these curve is $y = x + \sqrt 5 $
Statement $-2$: If the line, $y = mx + \frac{{\sqrt 5 }}{m}\left( {m \ne 0} \right)$ is their common tangent , then $m$ satisfies ${m^4} - 3{m^2} + 2 = 0$.