Area Question 1
Question 1
- 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) $4 \sqrt{3}$
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Solution:
- Given, equation of curves are
$$ \begin{aligned} y^{2} & =4 \lambda x \ y & =\lambda x \ \lambda & >0 \end{aligned} $$
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
$$ \begin{aligned} & =\int_{0}^{4 / \lambda}(2 \sqrt{\lambda x}-\lambda x) d x \ & =2 \sqrt{\lambda} \frac{x^{3 / 2}}{\frac{3}{2}} 0 \ & =\frac{4}{3} \sqrt{\lambda} \frac{4 \sqrt{4}}{\lambda \sqrt{\lambda}}-\frac{\lambda}{2} \frac{4}{\lambda}^{2} \ & =\frac{32}{3 \lambda}-\frac{8}{\lambda}=\frac{32-24}{3 \lambda}=\frac{8}{3 \lambda} \end{aligned} $$
It is given that area $=\frac{1}{9}$
