Q97 / 160 · AP EAMCET 2021 (23 Aug Shift 2) · Mathematics
If the point $(a, 2 a)$ is an interior point of the region bounded by the parabola $y^2=16 x$ and the double ordinate through focus, then
- A$a < 4$
- Correct answerB$0 < a < 4$
- C$0 < a < 2$
- D$a>4$
Solution
Given parabola,
$y^2=16 x$ ...(i)
Comparing with $y^2=4 p x$
$\therefore$ Focus $S=(4,0)$ $\because P(a, 2 a)$ is in interior region of the parabola $y^2-16 x=0$ $\therefore \quad(2 a)^2-16 a < 0$ $\Rightarrow \quad a^2-4 a < 0$ Also $(a, 2 a)$ and vertex $(0,0)$ are on the same side of $x=4$ $\begin{array}{lr}\Rightarrow & x-4=0 \\ \text { So, } & a-4 < 0 \\ \Rightarrow & a < 4\end{array}$ Now, $a^2-4 a < 0$ $\Rightarrow \quad 0 < a < 4$
$\therefore$ Focus $S=(4,0)$ $\because P(a, 2 a)$ is in interior region of the parabola $y^2-16 x=0$ $\therefore \quad(2 a)^2-16 a < 0$ $\Rightarrow \quad a^2-4 a < 0$ Also $(a, 2 a)$ and vertex $(0,0)$ are on the same side of $x=4$ $\begin{array}{lr}\Rightarrow & x-4=0 \\ \text { So, } & a-4 < 0 \\ \Rightarrow & a < 4\end{array}$ Now, $a^2-4 a < 0$ $\Rightarrow \quad 0 < a < 4$
Hence, the correct answer is option: 'B'.