Let T and C respectively be the transverse and conjugate axes of the hyperbola 16x^2 - y^2…

Question

Let T and C respectively be the transverse and conjugate axes of the hyperbola $16{x^2} - {y^2} + 64x + 4y + 44 = 0$. 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 :

Accepted Answer

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

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

TA $: y=2$

$\mathrm{CA}: x=-2$

$$
A=\left|\int_{-2}^{\sqrt{6}}\left(2-\left(x^{2}-4\right)\right) d x\right|
$$

$$
\begin{aligned}
& =6 x-\left.\frac{x^{3}}{3}\right|_{-2} ^{\sqrt{6}} \\
& =\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right)
\end{aligned}
$$

$$
=\frac{12 \sqrt{6}}{3}+\frac{28}{3}
$$

Solutions · 1