The ordinates of the points P and $\mathrm{Q}$ on the parabola with focus $(3,0)$ and directrix $x=-3$ are in the ratio $3: 1$. If $\mathrm{R}(\alpha, \beta)$ is the point of intersection of the tangents to the parabola at $\mathrm{P}$ and $\mathrm{Q}$, then $\frac{\beta^{2}}{\alpha}$ is equal to _______________.
Answer (integer)
16
Solution
$$
\begin{aligned}
& \text { Give parabola is : } y^2=12 x \quad(\because a=3) \\\\
& \text { So, } \mathrm{P} \equiv\left(a t_1^2, 2 a t_1\right) \\\\
& \mathrm{Q} \equiv\left(a t_2^2, 2 a t_2\right) \\\\
& \text { So, point } \mathrm{R}(\alpha, \beta) \equiv\left(a t_1 t_2, a\left(t_1+t_2\right)\right) \\\\
& \equiv((3 t)(3 t), 3(t+3 t))=\left(9 t^2, 12 t\right) \\\\
& \therefore \frac{\beta^2}{\alpha}=\frac{144 t^2}{9 t^2}=16
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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