The area of the smaller region enclosed by the curves y2=8 x+4 and x2+y2+4 √3 x-4=0 is equal to

Question

The area of the smaller region enclosed by the curves y2=8 x+4 and x2+y2+4 √3 x-4=0 is equal to (A) (1/3)(2-12 √3+8 π) (B) (1/3)(2-12 √3+6 π) (C) (1/3)(4-12 √3+8 π) (D) (1/3)(4-12 √3+6 π)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/950a3ca265ae139da9a9c9b79338914c-.png" /> x2+y2+4 √3 x-4=0 y2=8 x+4 Point of intersections are (0,2) &(0,-2) Both are symmetric about x-axis Area =2 ∫ limits02(√16-y2-2 √3)-((y2-4/8)) d y On solving Area =(1/3)[8 π+4-12 √3]

Q. The area of the smaller region enclosed by the curves $y^{2} = 8 x + 4$ and $x^{2} + y^{2} + 43 x - 4 = 0$ is equal to

$\frac{1}{3} (2 - 123 + 8 π)$

$\frac{1}{3} (2 - 123 + 6 π)$

$\frac{1}{3} (4 - 123 + 8 π)$

$\frac{1}{3} (4 - 123 + 6 π)$

$x^{2} + y^{2} + 43 x - 4 = 0$
$y^{2} = 8 x + 4$
Point of intersections are $(0, 2) & (0, - 2)$ Both are symmetric about $x$ -axis Area $= 2 0 \int 2 (16 - y^{2} - 23) - (\frac{y ^{2} - 4}{8}) d y$ On solving Area $= \frac{1}{3} [8 π + 4 - 123]$

Q. The area of the smaller region enclosed by the curves y2=8x+4 and x2+y2+43​x−4=0 is equal to

31​(2−123​+8π)

31​(2−123​+6π)

31​(4−123​+8π)

31​(4−123​+6π)