Question

The ratio in which the area bounded by the curves $y^2=12x$ and $x^2=12y$ is divided by the line x $=$ 3 is

• A. 15 : 16
• B. 15 : 49
• C. 1 : 2
• D. None of these

Answer 1

The correct option is B 15 : 49

$y^2=12x$ $x^2=12y$
${\int }_0^3\sqrt{12x}-{\int }_0^3\frac{x^2}{12}={\int }_0^3\sqrt{12x}-\frac{x^2}{12}$
$\frac{2\sqrt{12}{x}^{\frac{3}{2}}}{3}-\frac{x^3}{36}{\int }_0^3$
$\frac{\sqrt{12}3\sqrt{3}}{3}-\frac{27}{36}$
$12-\frac{3}{4}=\frac{45}{4}$
Same as $\frac{8}{4}$ for remaining part
$\therefore ratio=\frac{15}{49}$