SATHEE: Area Question 1

Area Question 1

Question 1

If the area (in sq units) bounded by the parabola $y^{2} = 4 λ x$ and the line $y = λ x, λ > 0$ , is $\frac{1}{9}$ , then $λ$ 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

$\begin{aligned} y^{2} & = 4 λ x y & = λ x λ & > 0 \end{aligned}$

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

$\begin{aligned} λ^{2} x^{2} = 4 λ x & \Rightarrow x = \frac{4}{λ}, so y = 4 . & So, A \frac{4}{λ}, 4 \end{aligned}$

Now, required area is

$\begin{aligned} = \int_{0}^{4 / λ} (2 \sqrt{λ x} - λ x) d x & = 2 \sqrt{λ} \frac{x^{3 / 2}}{\frac{3}{2}} 0 & = \frac{4}{3} \sqrt{λ} \frac{4 \sqrt{4}}{λ \sqrt{λ}} - \frac{λ}{2} {\frac{4}{λ}}^{2} & = \frac{32}{3 λ} - \frac{8}{λ} = \frac{32 - 24}{3 λ} = \frac{8}{3 λ} \end{aligned}$

It is given that area $= \frac{1}{9}$

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