JEE Mathematics 2023 Question with Solution

Given: hyperbola $16x^2 - y^2 + 64x + 4y + 44 = 0$, parabola $x^2 = y + 4$.

Find: area of the region above the parabola, below the transverse axis, and to the right of the conjugate axis.

Complete the square in the hyperbola equation:

$$16(x^2 + 4x) - (y^2 - 4y) + 44 = 0$$ $$16(x + 2)^2 - 64 - (y - 2)^2 + 4 + 44 = 0$$ $$16(x + 2)^2 - (y - 2)^2 = 16$$ $$\frac{(x + 2)^2}{1} - \frac{(y - 2)^2}{16} = 1$$

So the center is $(-2,2)$. Hence the transverse axis is $y = 2$ and the conjugate axis is $x = -2$.

The parabola is

$$y = x^2 - 4$$

Therefore the required region is bounded above by $y = 2$, below by $y = x^2 - 4$, and on the left by $x = -2$.

Find the right intersection of the parabola with $y = 2$:

$$x^2 - 4 = 2$$ $$x^2 = 6$$ $$x = \sqrt{6}$$

Thus the area is

$$A = \int_{-2}^{\sqrt{6}} \left(2 - (x^2 - 4)\right) \, dx$$ $$A = \int_{-2}^{\sqrt{6}} (6 - x^2) \, dx$$ $$A = \left[6x - \frac{x^3}{3}\right]_{-2}^{\sqrt{6}}$$

At $x = \sqrt{6}$:

$$6\sqrt{6} - \frac{(\sqrt{6})^3}{3} = 6\sqrt{6} - 2\sqrt{6} = 4\sqrt{6}$$

At $x = -2$:

$$6(-2) - \frac{(-2)^3}{3} = -12 + \frac{8}{3} = -\frac{28}{3}$$

Therefore,

$$A = 4\sqrt{6} - \left(-\frac{28}{3}\right) = 4\sqrt{6} + \frac{28}{3}$$

So the correct option is B.

The solution shows an inconsistent label "C", but the working and final value match option B.

From

$$\frac{(x + 2)^2}{1} - \frac{(y - 2)^2}{16} = 1$$

the hyperbola has center $(-2,2)$. For this hyperbola, the transverse axis is the horizontal line through the center, so $T: y = 2$. The conjugate axis is the vertical line through the center, so $C: x = -2$.

Since the region is above the parabola and below $T$, its vertical thickness at a point $x$ is

$$2 - (x^2 - 4) = 6 - x^2$$

The phrase "on the right of the conjugate axis" gives $x \ge -2$. The region ends where the parabola meets $y = 2$, giving $x = \sqrt{6}$ on the right side. Hence the same integral follows:

$$\int_{-2}^{\sqrt{6}} (6 - x^2) \, dx$$

which evaluates to

$$4\sqrt{6} + \frac{28}{3}$$