Area Ques 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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Answer:
Correct Answer: 1.(c)
Solution: (c) Given, equation of curves are
and
$y^2 =4 \lambda x $ $\quad$ ……..(i)
$y =\lambda x $ $\quad$ ……..(ii)
$\lambda >0$
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
$\Rightarrow \lambda^2 x^2=4 \lambda x $
$ x=\frac{4}{\lambda}, \text { so } y=4 $
So, $A\left(\frac{4}{\lambda}, 4\right)$
Now, required area is
$ \begin{aligned} & =\int_0^{4 \lambda}(2 \sqrt{\lambda x}-\lambda x) d x \\ & =2 \sqrt{\lambda}\left[\frac{x^{3 / 2}}{\frac{3}{2}}\right]_0^{4 / \lambda}-\lambda\left[\frac{x^2}{2}\right]_0^{4 \lambda} \\ & =\frac{4}{3} \sqrt{\lambda} \frac{4 \sqrt{4}}{\lambda \sqrt{\lambda}}-\frac{\lambda}{2}\left(\frac{4}{\lambda}\right)^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 \frac{8}{3 \lambda}=\frac{1}{9} $
$\Rightarrow \lambda =24$
BETA
