1e:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1f:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1d:T42d,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"The sum of squares of all possible values of k for which are","item":"https://www.divinejee.com/question/the-sum-of-squares-of-all-possible-values-of-k-for-which-area-of-the-region-bounded-by-the"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question The sum of squares of all possible values of k for which are on Topic","text":"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)...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/the-sum-of-squares-of-all-possible-values-of-k-for-which-area-of-the-region-bounded-by-the"}}}]}20:T4f0,Given $k y^2=2(y-x)$ .........(i)

$$ 2 y^2=k x $$ .........(ii)

Point of intersection of (i) and (ii)

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

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

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

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

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

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

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