The sum of squares of all possible values of k, for which area of the region bounded by th…

Question

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 :

Answer: 8

Accepted Answer

Given $k{y}^{2}=2(y-x)$ .........(i)

$2{y}^{2}=kx$ .........(ii)

Point of intersection of (i) and (ii)

$\begin{gathered} k{y}^{2}=2(y-\frac{2{y}^{2}}{k}) \ \Rightarrow y=0,ky=2(1-\frac{2y}{k}) \end{gathered}$

$\begin{gathered} ky+\frac{4y}{k}=2 \ y=\frac{2}{k+\frac{4}{k}}=\frac{2k}{{k}^{2}+4} \end{gathered}$

$\begin{gathered} A=\int _{6}^{\frac{2k}{{k}^{2}+4}}((y-\frac{k{y}^{2}}{2})-\frac{2{y}^{2}}{k})dy \ A={[\frac{{y}^{2}}{2}-(\frac{k}{2}+\frac{2}{k})\frac{{y}^{3}}{3}]}_{\frac{2k}{{k}^{2}+4}}^{0} \end{gathered}$

$\begin{gathered} ={(\frac{2k}{{k}^{2}+4})}^{2}[\frac{1}{2}-\frac{{k}^{2}+4}{2k}(\frac{1}{3})(\frac{2k}{{k}^{2}+4})] \ =\frac{1}{6}	imes 4	imes {(\frac{1}{k+\frac{4}{k}})}^{2} \end{gathered}$

$\begin{gathered} \ge \ \frac{(k+\frac{4}{k})}{2}\ge 2 \ k+\frac{4}{k}\ge 4 \end{gathered}$

$	herefore $ Area is maximum when $k=\frac{4}{k}$

$\begin{gathered} 	herefore {k}^{2}=4 \ k=\pm 2 \ {k}_{1}=2,{k}_{2}=-2 \ 	herefore {k}_{2}^{1}+{k}_{2}^{2}=(+2{)}^{2}+(-2{)}^{2} \ =4+4 \ =08 \end{gathered}$

Solutions · 1