The sum of squares of all possible values of k, for which area of the region bounded by the parabolas 2 y 2 = k x and k y 2 = 2 ( y - x ) is maximum, is equal to [2024]

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Solution

Q.
The sum of squares of all possible values of $k,$ for which area of the region bounded by the parabolas $2 y^{2} = k x$ and $k y^{2} = 2 (y - x)$ is maximum, is equal to _______ . [2024]

Ans.
(8)

$k y^{2} = 2 (y - x)$ and $2 y^{2} = k x$

Point of intersection

$y (k y - 2 (1 - \frac{2 y}{k})) = 0$

$\Rightarrow k y^{2} - 2 (y - \frac{2 y^{2}}{k}) = 0$

$\Rightarrow y = 0 and k y = 2 (1 - \frac{2 y}{k}) \Rightarrow k y + \frac{4 y}{k} = 2$

$y = \frac{2}{k + \frac{4}{k}} = \frac{2 k}{k^{2} + 4}$ $∴ A = \int_{0}^{\frac{2 k}{k^{2} + 4}} ((y - \frac{k y^{2}}{2}) - (\frac{2 y^{2}}{k})) \cdot d y$

$= {[\frac{y^{2}}{2} - (\frac{k}{2} + \frac{2}{k}) \cdot \frac{y^{3}}{3}]}_{0}^{\frac{2 k}{k^{2} + 4}}$

$= {(\frac{2 k}{k^{2} + 4})}^{2} [\frac{1}{2} - \frac{k^{2} + 4}{2 k} \times \frac{1}{3} \times \frac{2 k}{k^{2} + 4}] = \frac{1}{6} \times 4 \times {(\frac{1}{k + \frac{4}{k}})}^{2}$

$A.M. \geq G.M. \Rightarrow (\frac{k + \frac{4}{k}}{2}) \geq 2 \Rightarrow k + \frac{4}{k} \geq 4$

$So, area is maximum when k = \frac{4}{k} \Rightarrow k = 2, - 2$

$∴ Sum of squares of all possible values of k$

$= 2^{2} + {(- 2)}^{2} = 8$

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