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 $2 \mathrm{y}^{2}=\mathrm{kx}$ and $\mathrm{ky}^{2}=2(\mathrm{y}-\mathrm{x})$ is maximum, is equal to :
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Answer (8)
Solution
$k y^{2}=2(y-x)$
$2 \mathrm{y}^{2}=\mathrm{kx}$
Point of intersection $\rightarrow$
$\mathrm{ky}^{2}=\left(\frac{\mathrm{y}-2 \mathrm{y}^{2}}{\mathrm{k}}\right)$
$\mathrm{y}=0 \quad \mathrm{ky}=2\left(\frac{1-2 \mathrm{y}}{\mathrm{k}}\right)$
$\mathrm{ky}+\frac{4 \mathrm{y}}{\mathrm{k}}=2$
$\mathrm{y}=\frac{2}{\mathrm{k}+\frac{4}{\mathrm{k}}}=\frac{2 \mathrm{k}}{\mathrm{k}^{2}+4}$
$A=\int_{0}^{\frac{2 k}{k^{2}+4}}\left(\left(y-\frac{k y^{2}}{2}\right)-\left(\frac{2 y^{2}}{k}\right)\right) \cdot d y$
$=\frac{y^{2}}{2}-\left.\left(\frac{k}{2}+\frac{2}{k}\right) \cdot \frac{y^{3}}{3}\right|_{0} ^{\frac{2 k}{k^{2}+4}}$
$=\left(\frac{2 k}{k^{2}+4}\right)^{2}\left[\frac{1}{2}-\frac{k^{2}+4}{2 k} \times \frac{1}{3} \times \frac{2 k}{k^{2}+4}\right]$
$=\frac{1}{6} \times 4 \times\left(\frac{1}{k+\frac{4}{k}}\right)^{2}$
$A \cdot M \geq G \cdot M \frac{\left(k+\frac{4}{k}\right)}{2} \geq 2$
$\mathrm{k}+\frac{4}{\mathrm{k}} \geq 4$
Area is maximum when $\mathrm{k}=\frac{4}{\mathrm{k}}$
$\mathrm{k}=2,-2$
