The sum of squares of all possible values of | vedclass

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(D) The given parabolas are $2y^2 = kx$ and $ky^2 = 2(y - x)$.To find the intersection points,substitute $x = \frac{2y^2}{k}$ into the second equation

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) The given parabolas are $2y^2 = kx$ and $ky^2 = 2(y - x)$.To find the intersection points,substitute $x = \frac{2y^2}{k}$ into the second equation:$ky^2 = 2(y - \frac{2y^2}{k})$$ky^2 = 2y - \frac{4y^2}{k}$$y^2(k + \frac{4}{k}) = 2y$$y(y(k + \frac{4}{k}) - 2) = 0$So,$y = 0$ or $y = \frac{2}{k + \frac{4}{k}} = \frac{2k}{k^2 + 4}$.The area $A$ is given by:$A = \int_0^{\frac{2k}{k^2 + 4}} (x_2 - x_1) dy = \int_0^{\frac{2k}{k^2 + 4}} ((y - \frac{ky^2}{2}) - \frac{2y^2}{k}) dy$$A = \int_0^{\frac{2k}{k^2 + 4}} (y - (\frac{k}{2} + \frac{2}{k})y^2) dy$$A = [\frac{y^2}{2} - (\frac{k^2 + 4}{2k}) \frac{y^3}{3}]_0^{\frac{2k}{k^2 + 4}}$$A = \frac{1}{2}(\frac{2k}{k^2 + 4})^2 - \frac{k^2 + 4}{6k} (\frac{2k}{k^2 + 4})^3 = \frac{1}{6} (\frac{2k}{k^2 + 4})^2 = \frac{2}{3} (\frac{k}{k^2 + 4})^2 = \frac{2}{3} \frac{1}{(k + \frac{4}{k})^2}$.For the area to be maximum,the denominator $(k + \frac{4}{k})^2$ must be minimum.By $AM \geq GM$,$k + \frac{4}{k} \geq 2\sqrt{k \cdot \frac{4}{k}} = 4$ (for $k > 0$) or $k + \frac{4}{k} \leq -4$ (for $k The minimum value of $(k + \frac{4}{k})^2$ is $16$,which occurs when $k = \frac{4}{k}$,i.e.,$k^2 = 4$,so $k = 2$ or $k = -2$.The sum of squares of these values is $2^2 + (-2)^2 = 4 + 4 = 8$.

The sum of squares of all possible values of $k$,for which the area of the region bounded by the parabolas $2y^2 = kx$ and $ky^2 = 2(y - x)$ is maximum,is equal to:

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