Q. If $(9a,\ 6a)$ is a point bounded in region formed by parabola $y^{2} = 16x$ and $x = 9$, then
A) $a \in (0,1)$
B) $a < \frac{1}{4}$
C) $a < 1$
D) $0 < a < 4$
Explanation
(a)
Since, the point $(9a,\ 6a)$ is bounded in the region formed by the parabola $y^{2} = 16x$ and $x = 9$, then
$y^{2} - 16x < 0,\ x - 9 < 0$
$\Rightarrow {36a}^{2} - 16 \bullet 9a < 0,\ 9a - 9 < 0$
$\Rightarrow 36a(a - 4) < 0,\ a < 1$
$0 < a < 4,\ a < 1\ \Rightarrow 0 < a < 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.