JEE Main 2024 — Mathematics Calculus

Given: $k{y}^{2}=2(y-x),2{y}^{2}=kx$ky2=2(y−x),2y2=kx

Finding the point of intersection,

$$\Rightarrow k{y}^{2}=2(y-\frac{2{y}^{2}}{k})$$⇒ky2=2(y−k2y2​)

$\Rightarrow y=0$⇒y=0 and $ky=2(1-\frac{2y}{k})$ky=2(1−k2y​)

$$\Rightarrow ky+\frac{4y}{k}=2$$⇒ky+k4y​=2

$$\Rightarrow y=\frac{2}{k+\frac{4}{k}}$$⇒y=k+k4​2​

$$\Rightarrow y=\frac{2k}{{k}^{2}+4}$$⇒y=k2+42k​

So, the required area is given by,

$$A=\int _{0}^{\frac{2k}{{k}^{2}+4}}((y-\frac{k{y}^{2}}{2})-(\frac{2{y}^{2}}{k}))\cdot dy$$A=∫0k2+42k​​((y−2ky2​)−(k2y2​))⋅dy

$$\Rightarrow A={[\frac{{y}^{2}}{2}-(\frac{k}{2}+\frac{2}{k})\cdot \frac{{y}^{3}}{3}]}_{0}^{\frac{2k}{{k}^{2}+4}}$$⇒A=[2y2​−(2k​+k2​)⋅3y3​]0k2+42k​​

$$\Rightarrow A={(\frac{2k}{{k}^{2}+4})}^{2}[\frac{1}{2}-\frac{{k}^{2}+4}{2k}\times \frac{1}{3}\times \frac{2k}{{k}^{2}+4}]$$⇒A=(k2+42k​)2[21​−2kk2+4​×31​×k2+42k​]

$$\Rightarrow A=\frac{1}{6}\times 4\times {(\frac{1}{k+\frac{4}{k}})}^{2}$$⇒A=61​×4×(k+k4​1​)2

We know that, $AM\geq GM$AM≥GM

$$\Rightarrow \frac{(k+\frac{4}{k})}{2}\geq 2$$⇒2(k+k4​)​≥2

$$\Rightarrow k+\frac{4}{k}\geq 4$$⇒k+k4​≥4

So, area is maximum when $k=\frac{4}{k}$k=k4​.

$$\Rightarrow k=2,-2$$⇒k=2,−2

Hence, the sum of squares of all possible values of $k$k is $8$8.