JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 8)
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 :
$$4\sqrt 6 - {{28} \over 3}$$
$$4\sqrt 6 - {{44} \over 3}$$
$$4\sqrt 6 + {{28} \over 3}$$
$$4\sqrt 6 + {{44} \over 3}$$
Explanation
$16(x+2)^{2}-(y-2)^{2}=16$
$$ \frac{(x+2)^{2}}{1}-\frac{(y-2)^{2}}{16}=1 $$
TA $: y=2$
$\mathrm{CA}: x=-2$
_25th_January_Evening_Shift_en_8_1.png)
$$ A=\left|\int_{-2}^{\sqrt{6}}\left(2-\left(x^{2}-4\right)\right) d x\right| $$
$$ \begin{aligned} & =6 x-\left.\frac{x^{3}}{3}\right|_{-2} ^{\sqrt{6}} \\\\ & =\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right) \end{aligned} $$
$$ =\frac{12 \sqrt{6}}{3}+\frac{28}{3} $$
$$ \frac{(x+2)^{2}}{1}-\frac{(y-2)^{2}}{16}=1 $$
TA $: y=2$
$\mathrm{CA}: x=-2$
$$ A=\left|\int_{-2}^{\sqrt{6}}\left(2-\left(x^{2}-4\right)\right) d x\right| $$
$$ \begin{aligned} & =6 x-\left.\frac{x^{3}}{3}\right|_{-2} ^{\sqrt{6}} \\\\ & =\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right) \end{aligned} $$
$$ =\frac{12 \sqrt{6}}{3}+\frac{28}{3} $$
Comments (0)
