Q. The ends of a line segment are $P(1,\ 3)$ and $Q(1,\ 1).$ $R$ is a point on the line segment $PQ$ such that $PR:QR = 1:\lambda$. If $R$ is an interior point of the parabola $y^{2} = 4x$, then
A) $\lambda \in (0,\ 1)$
B) $\lambda \in \left( - \frac{3}{5},\ 1 \right)$
C) $\lambda \in \left( \frac{1}{2},\ \frac{3}{5} \right)$
D) None of these
Explanation
(a)
Let any point on the line segment $PQ$ is $R(\alpha,\ \beta)$, then
$\alpha = \frac{\lambda(1) + 1}{\lambda + 1} = 1$,
And $\beta = \frac{3\lambda + 1}{\lambda + 1}$ $(\because\lambda > 0\ as\ R\ is\ on\ segment\ AB)$
A point is inside parabola $y^{2} = 4x$, if
$y^{2} - 4x < 0$
$\Rightarrow \left( \frac{3\lambda + 1}{\lambda + 1} \right)^{2} - 4(1) < 0$
$\Rightarrow \left( \frac{3\lambda + 1}{\lambda + 1} + 2 \right)\left( \frac{3\lambda + 1}{\lambda + 1} - 2 \right) < 0$
$\Rightarrow (5\lambda + 3)(\lambda - 1) < 0$
$\Rightarrow - \frac{3}{5} < \lambda < 1$
So, $0 < \lambda < 1$ (but $\lambda > 0$)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.