JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If the area enclosed by the parabola $P_{1} : 2 y = 5 x^{2}$ and $P_{2} : x^{2} - y + 6 = 0$ is equal to the area enclosed by $P_{1}$ and $y = \alpha x, \alpha > 0$ , then $α^{3}$ is equal to:

Choose the correct answer:

Explanation

Solution

Step 1: Find Area $A_1$ between $P_1$ and $P_2$

Intersection points:

\frac{5}{2}x^2 = x^2 + 6

\frac{3}{2}x^2 = 6 \implies x^2 = 4 \implies x = \pm 2

A_1 = \int_{-2}^{2} \left( (x^2 + 6) - \frac{5}{2}x^2 \right) dx

A_1 = \int_{-2}^{2} \left( 6 - \frac{3}{2}x^2 \right) dx

A_1 = \left[ 6x - \frac{x^3}{2} \right]_{-2}^{2}

A_1 = (12 - 4) - (-12 + 4) = 8 + 8 = 16

Step 2: Find Area $A_2$ between $P_1$ and $y = \alpha x$

Intersection points:

\frac{5}{2}x^2 = \alpha x \implies x = 0, x = \frac{2\alpha}{5}

A_2 = \int_{0}^{2\alpha/5} \left( \alpha x - \frac{5}{2}x^2 \right) dx

A_2 = \left[ \frac{\alpha x^2}{2} - \frac{5x^3}{6} \right]_{0}^{2\alpha/5}

A_2 = \frac{\alpha}{2}\left(\frac{4\alpha^2}{25}\right) - \frac{5}{6}\left(\frac{8\alpha^3}{125}\right)

A_2 = \frac{2\alpha^3}{25} - \frac{4\alpha^3}{75} = \frac{6\alpha^3 - 4\alpha^3}{75} = \frac{2\alpha^3}{75}

Step 3: Equate $A_1$ and $A_2$

16 = \frac{2\alpha^3}{75}

8 = \frac{\alpha^3}{75}

\alpha^3 = 8 \times 75

\alpha^3 = 600

Final Answer:

\alpha^3 = 600