JEE Mathematics 2023 Question with Solution

Given: The parabolas are $P_1: 2y = 5x^2$ and $P_2: x^2 - y + 6 = 0$. Also, the area enclosed by these two parabolas is equal to the area enclosed by $P_1$ and $y = \alpha x$.

Find: $\alpha^3$.

From $P_1$, we get

and from $P_2$, we get

The point of intersection satisfies

so the abscissae of intersection are $x = \pm 2$.

Hence the enclosed area between the two parabolas is

This is equal to the area enclosed by $P_1$ and $y = \alpha x$, that is

Therefore,

Using the result shown in the solution,

so

Therefore, the required value is $600$.

Given: Equal enclosed areas are formed by $P_1$ with $P_2$ and by $P_1$ with the line $y = \alpha x$.

Find: The value of $\alpha^3$.

For the first region, the upper curve is $y = x^2 + 6$ and the lower curve is $y = \frac{5}{2}x^2$. Their intersections occur at $x = -2$ and $x = 2$, so symmetry gives

the solution evaluates this enclosed area as $16$.

For the second region, the line $y = \alpha x$ intersects $y = \frac{5}{2}x^2$ at

which gives the non-zero intersection point

Thus the area is

Equating this to $16$ leads to the final result