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. $15:49$
• B. $49:15$
• C. $2:3$
• D. $3:2$

Answer 1

The correct option is A $15:49$

Given curves are $y^2=12x$ and $x^2=12y$ and $x=3$


Here, $A_1={\int }_0^3(2\sqrt{3}\sqrt{x}-\frac{x^2}{12})dx$

$={(2\sqrt{3}\left(\frac{2}{3}{x}^{3/2}\right)-\frac{x^3}{36})}_0^3$

$=2\sqrt{3}(\frac{2}{3}\times 3\sqrt{3})-\frac{27}{36}=12-\frac{3}{4}=\frac{45}{4}\text{sq.units}$

Also,

$A_2={\int }_3^{12}(2\sqrt{3}\sqrt{x}-\frac{x^2}{12})dx={(2\sqrt{3}\left(\frac{2}{3}{x}^{3/2}\right)-\frac{x^3}{36})}_3^{12}=(2\sqrt{3}\times \frac{2}{3}\times 12\sqrt{12}-\frac{{12}^3}{36})-(\frac{2}{3}\times 2\sqrt{3}\times 3\sqrt{3}-\frac{27}{36})=(96-48)-(12-\frac{3}{4})=48-12+\frac{3}{4}=36+\frac{3}{4}=\frac{147}{4}\text{sq.units}$

Thus, $A_1:A_2=45:147$

$=15:49$