CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image