ParabolaHardBloom L3
Question
In a parabola $y^{2} = 4ax$ the angle $\theta$ that the latus rectum subtends at the vertex of the parabola is:
Options
A.dependent on the length of the latus rectum
B.independent of the latus rectum and lies between $\frac{5\pi}{6}$ and $\pi$
C.independent of the latus rectum and lies between $\frac{3\pi}{4}$ and $\frac{5\pi}{6}$
D.independent of the latus rectum and lies between $\frac{2\pi}{3}$ and $\frac{3\pi}{4}$
Solution
Sol. [Key Concept: The latus-rectum ends are $(a, \pm \ 2a)$; the angle they subtend at the vertex is $2\tan^{- 1}2$, independent of $a$ — then locate its interval.]
$$\theta = 2\tan^{- 1}2 $$
$$\begin{matrix} & \sqrt{3} < 2 < \sqrt{2} + 1 \\ \Rightarrow & \frac{\pi}{3} < \tan^{- 1}2 < \frac{3\pi}{8} \\ \Rightarrow & \frac{2\pi}{3} < \theta < \frac{3\pi}{4} \end{matrix}$$
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