Question

The ends of a line segment are $P(1,3)$ and $Q(1,1)$. $R$ is a point on line segment $PQ$ such that $PR:QR=1:\lambda$. If $R$ is an interior point of parabola $y^2=4x$, then

• A. $\lambda \in (0,1)$
• B. $\lambda \in (-\frac{3}{5},1)$
• C. $\lambda \in (\frac{1}{2},\frac{3}{5})$
• D. none of these

Answer 1

Answer: (A)

$\lambda \in (0,1)$

Let the coordinates of $R$ be $(\alpha ,\beta )$.

$\alpha =\frac{\lambda \left(1\right)+1}{\lambda +1}=1$, $\beta =\frac{3\lambda +1}{\lambda +1}$

For $R$ to be an interior point of the given parabola $y^2=4x$,

${\beta }^2-4\alpha <0$

$\Rightarrow {\left(\frac{3\lambda +1}{\lambda +1}\right)}^2-4<0$

$\Rightarrow (\frac{3\lambda +1}{\lambda +1}+2)(\frac{3\lambda +1}{\lambda +1}-2)<0$

$\Rightarrow (5\lambda +3)(\lambda -1)<0$

$\Rightarrow -\frac{3}{5}<\lambda <1$

But $\lambda >0$ ( $\because R$ is a point on the line segment $PQ$)

Hence, $\lambda \in (0,1)$

Hence, option A.