Let T and C respectively be the transverse and conjugate axes of the hyperbola 16 x 2 - y 2 + 64 x + 4 y + 44 = 0 . Then the area of the region above the parabola x 2 = y + 4 , below the transverse axis T and on the right of the conjugate axis C is:

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Solution

Q.
Let T and C respectively be the transverse and conjugate axes of the hyperbola $16 x^{2} - y^{2} + 64 x + 4 y + 44 = 0$ . Then the area of the region above the parabola $x^{2} = y + 4,$ below the transverse axis T and on the right of the conjugate axis C is: [2023]

1 $4 \sqrt{6} + \frac{44}{3}$

2 $4 \sqrt{6} - \frac{28}{3}$

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

4 $4 \sqrt{6} - \frac{44}{3}$

Ans.
(3)

$We have given 16 x^{2} - y^{2} + 64 x + 4 y + 44 = 0$

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

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

$16 {(x + 2)}^{2} - 64 - {(y - 2)}^{2} + 48 = 0$

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

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

$\Rightarrow Centre: (- 2, 2)$

$Vertices: (- 1, 2), (- 3, 2)$

$Equation of parabola: x^{2} = y + 4$

$Required area = \int_{- 2}^{\sqrt{6}} {2 - (x^{2} - 4)} d x$

$= \int_{- 2}^{\sqrt{6}} (6 - x^{2}) d x = {[6 x - \frac{x^{3}}{3}]}_{- 2}^{\sqrt{6}}$

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

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