Question:
If the area enclosed between the curves $y=k x^{2}$ and $x=\mathrm{ky}^{2},(\mathrm{k}>0)$, is 1 square unit. Then $\mathrm{k}$ is:
Correct Option: 1
Solution:
Area bounded by $y^{2}=4 a x \& x^{2}=4 b y, a, b \neq 0$
is $\left|\frac{16 a b}{3}\right|$
by using formula : $4 a=\frac{1}{k}=4 b, k>0$
Area $=\left|\frac{16 \cdot \frac{1}{4 k} \cdot \frac{1}{4 k}}{3}\right|=1$
$\Rightarrow k^{2}=\frac{1}{3}$
$\Rightarrow \mathrm{k}=\frac{1}{\sqrt{3}}$