QJEEM24Mad9e9: JEE Main 2024 Mathematics - Class 12 JEEM24Mad9e9 Solution: The sum of squares of all possible values of , for which area…

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The sum of squares of all possible values of

k

, for which area of the region bounded by the parabolas

2 y^2=\mathrm{k} x

and

\mathrm{ky}^2=2(y-x)

is maximum, is equal to :

Accepted Answer

Given

k y^2=2(y-x)

.........(i)

2 y^2=k x

.........(ii)

Point of intersection of (i) and (ii)

\begin{aligned}&amp; k y^2=2\left(y-\frac{2 y^2}{k}\right) \\\\&amp; \Rightarrow y=0, k y=2\left(1-\frac{2 y}{k}\right)\end{aligned}

\begin{aligned} &amp; k y+\frac{4 y}{k}=2 \\\\ &amp; y=\frac{2}{k+\frac{4}{k}}=\frac{2 k}{k^2+4}\end{aligned}

\begin{aligned} &amp; A=\int\limits_6^{\frac{2 k}{k^2+4}}\left(\left(y-\frac{k y^2}{2}\right)-\frac{2 y^2}{k}\right) d y \\\\ &amp; A=\left[\frac{y^2}{2}-\left(\frac{k}{2}+\frac{2}{k}\right) \frac{y^3}{3}\right]_0^{\frac{2 k}{k^2+4}}\end{aligned}

\begin{aligned} &amp; =\left(\frac{2 k}{k^2+4}\right)^2\left[\frac{1}{2}-\frac{k^2+4}{2 k}\left(\frac{1}{3}\right)\left(\frac{2 k}{k^2+4}\right)\right] \\\\ &amp; =\frac{1}{6} \times 4 \times\left(\frac{1}{k+\frac{4}{k}}\right)^2\end{aligned}

\begin{aligned} &amp; \text { A.M. } \geq \text { G.M. } \\\\ &amp; \frac{\left(k+\frac{4}{k}\right)}{2} \geq 2 \\\\ &amp; k+\frac{4}{k} \geq 4\end{aligned}

\therefore

Area is maximum when

k=\frac{4}{k}

\begin{aligned}&amp; \therefore k^2=4 \\\\&amp; k= \pm 2 \\\\&amp; k_1=2, k_2=-2 \\\\&amp; \therefore k_1^2+k_2^2=(+2)^2+(-2)^2 \\\\&amp; =4+4 \\\\&amp; =08\end{aligned}

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