JEE Main 2023 — Mathematics Calculus

Given:

$${P}_{1}:2y=5{x}^{2}$$P1​:2y=5x2

$${P}_{2}:{x}^{2}-y+6=0$$P2​:x2−y+6=0

On solving both equations, we have

$$2y=5(y-6)$$2y=5(y−6)

$$\Rightarrow 3y=30$$⇒3y=30

$$\Rightarrow y=10$$⇒y=10

$$\Rightarrow x=\pm 2$$⇒x=±2

So, points of intersection are (2,10)&(-2,10).

Also, solving ${P}_{1}$P1​ and $y=\alpha x$y=αx, we have

$$x=0,\frac{2\alpha }{5}$$x=0,52α​

According to question $\displaystyle 2{\int }_{0}^{2}({x}^{2}+6-\frac{5{x}^{2}}{2})dx={\int }_{0}^{\frac{2\alpha }{5}}(\alpha x-\frac{5{x}^{2}}{2})dx$2∫02​(x2+6−25x2​)dx=∫052α​​(αx−25x2​)dx

$$\Rightarrow 2{[\frac{{x}^{3}}{3}+6x-\frac{5{x}^{3}}{6}]}_{0}^{2}={[\frac{\alpha {x}^{2}}{2}-\frac{5{x}^{3}}{6}]}_{0}^{\frac{2\alpha }{5}}$$⇒2[3x3​+6x−65x3​]02​=[2αx2​−65x3​]052α​​

$$\Rightarrow 2(12+\frac{8}{3}-\frac{20}{3})=[\frac{2{\alpha }^{3}}{25}-\frac{8{\alpha }^{3}}{6\times 25}]$$⇒2(12+38​−320​)=[252α3​−6×258α3​]

$$\Rightarrow \frac{2{\alpha }^{3}}{75}=16$$⇒752α3​=16

$$\Rightarrow {\alpha }^{3}=600$$⇒α3=600