Area Ques 17
Question
- If the area (in sq units) bounded by the parabola $y^{2}=4 \lambda x$ and the line $y=\lambda x, \lambda>0$, is $\frac{1}{9}$, then $\lambda$ is equal to
(2019 Main, 12 April II)
(a) $2 \sqrt{6}$
(b) $48$
(c) $24$
(d) $4 \sqrt{3}$
Show Answer
Answer:
Correct Answer: 17.(c)
Solution:
Formula:
- Given, equation of curves are
$ \begin{aligned} y^{2} & =4 \lambda x \quad …….(i) \\ y & =\lambda x \quad …….(ii) \\ \lambda & >0 \end{aligned} $
Area bounded by above two curve is, as per figure
the intersection point $A$ we will get on the solving Eqs. (i) and (ii), we get
$ \begin{aligned} & \lambda^{2} x^{2}=4 \lambda x \\ & \Rightarrow \quad x=\frac{4}{\lambda} \text {, so } y=4 \text {. } \\ & \text { So, } \quad A (\frac{4}{\lambda}, 4) \end{aligned} $
Now, required area is
$ \begin{aligned} & =\int _{0}^{4 / \lambda}(2 \sqrt{\lambda x}-\lambda x) d x \\ & =2 \sqrt{\lambda} [\frac{x^{3 / 2}}{\frac{3}{2}}]_0^{4/\lambda}- \lambda [\frac{x^2}{2}]_0^{4/\lambda}\\ & =\frac{4}{3} \sqrt{\lambda} \frac{4 \sqrt{4}}{\lambda \sqrt{\lambda}}-\frac{\lambda}{2} (\frac{4}{\lambda})^{2} \\ & =\frac{32}{3 \lambda}-\frac{8}{\lambda}=\frac{32-24}{3 \lambda}=\frac{8}{3 \lambda} \end{aligned} $
It is given that area $=\frac{1}{9}$
$\Rightarrow \quad \frac{8}{\lambda}=\frac{1}{9}$
$\lambda = 24$
BETA
