Area Under Curves Question 3
Question 3 - 2024 (01 Feb Shift 2)
The sum of squares of all possible values of $k$, for which area of the region bounded by the parabolas $2 y^{2}=kx$ and $ky^{2}=2(y-x)$ is maximum, is equal to :
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Answer (8)
Solution
$k y^{2}=2(y-x)$
$2 y^{2}=kx$
Point of intersection $\rightarrow$
$ky^{2}=\left(\frac{y-2 y^{2}}{k}\right)$
$y=0 \quad ky=2\left(\frac{1-2 y}{k}\right)$
$ky+\frac{4 y}{k}=2$
$y=\frac{2}{k+\frac{4}{k}}=\frac{2 k}{k^{2}+4}$
$A=\int _0^{\frac{2 k}{k^{2}+4}}\left(\left(y-\frac{k y^{2}}{2}\right)-\left(\frac{2 y^{2}}{k}\right)\right) \cdot d y$
$=\frac{y^{2}}{2}-\left.\left(\frac{k}{2}+\frac{2}{k}\right) \cdot \frac{y^{3}}{3}\right| _0 ^{\frac{2 k}{k^{2}+4}}$
$=\left(\frac{2 k}{k^{2}+4}\right)^{2}\left[\frac{1}{2}-\frac{k^{2}+4}{2 k} \times \frac{1}{3} \times \frac{2 k}{k^{2}+4}\right]$
$=\frac{1}{6} \times 4 \times\left(\frac{1}{k+\frac{4}{k}}\right)^{2}$
$A \cdot M \geq G \cdot M \frac{\left(k+\frac{4}{k}\right)}{2} \geq 2$
$k+\frac{4}{k} \geq 4$
Area is maximum when $k=\frac{4}{k}$
$k=2,-2$
