Question

# 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 the interior point of the parabola $$y^2=4x$$, then :

A
λ(0,1)
B
λ(3,15)
C
λ(23,25)
D
λ(1,15)

Solution

## The correct option is A $$\lambda \in (0, 1)$$Let the coordinates of $$R$$ be $$(\alpha, \beta)$$.$$\alpha = \dfrac{ \lambda (1)+1}{ \lambda +1}=1$$, $$\beta = \dfrac{3 \lambda +1}{ \lambda+1}$$$$R$$ is an interior point of the given parabola $$y^2 =4x$$,$$\implies \beta^2 -4 \alpha < 0$$$$\implies \left( \dfrac{3 \lambda +1}{ \lambda+1} \right)^2 - 4 < 0$$$$\implies \left( \dfrac{3 \lambda +1}{ \lambda+1} +2 \right) \left( \dfrac{3 \lambda +1}{ \lambda+1} -2 \right) < 0$$$$\implies ( 5 \lambda +3)( \lambda -1) < 0$$$$\implies - \dfrac{3}{5} < \lambda < 1$$But $$\lambda > 0$$ ( $$\because R$$ is a point on the line segment $$PQ$$)Hence, $$\lambda \in (0,1)$$Hence, option A.Mathematics

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