ParabolaHardBloom L3
Question
If the point $\left( sin\theta,\frac{1}{\sqrt{2}} \right)$ lies exterior to both the parabolas $y^{2} = |x|$, then $\theta$ can belong to :
Options
A.$\left( 0,\frac{\pi}{6} \right)$
B.$\left( - \frac{\pi}{6},0 \right)$
C.$\left( \frac{5\pi}{6},\frac{7\pi}{6} \right)$
D.$\left( \frac{\pi}{6},\frac{5\pi}{6} \right)$
Solution
Sol. [Key Concept: A point is exterior to $y^{2} = |x|$ when $S_{1} > 0$ on both branches; impose $(\frac{1}{2})^{2} > |\sin\theta|$ to get the $\theta$-range.]
For the point $(sin\theta,\frac{1}{\sqrt{2}})$ to be exterior to both $y^{2} = x$ and $y^{2} = - x$, require $S_{1} > 0$ on each branch, which reduces to $|\sin\theta| < \frac{1}{2}$.
Hence $\theta \in (0,\frac{\pi}{6}) \cup ( - \frac{\pi}{6},0) \cup (\frac{5\pi}{6},\frac{7\pi}{6})$ → (a), (b), (c).
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