Question

The ratio of the area's bounded by the curves $y^2=12x$ and $x^2=12y$ is divided by the line $x=3$ is

• A. 15 : 49
• B. 9 : 15
• C. 7 : 15
• D. 7 : 5

Answer 1

The correct option is A 15 : 49

For the point of intersection
$2\sqrt{3x}-\frac{x^2}{12}=0$
Or
$\sqrt{x}[24\sqrt{3}-x\sqrt{x}]=0$
$x=0$ and $x=12$.
Hence
${\int }_0^3(2\sqrt{3x}-\frac{x^2}{12}).dx$
$=[\frac{4\sqrt{3}}{3}.x^{\frac{3}{2}}-\frac{x^3}{36}{]}_0^3$
$=\frac{45}{4}$ ...(i)
And
${\int }_3^{12}(2\sqrt{3x}-\frac{x^2}{12}).dx$
$=[\frac{4\sqrt{3}}{3}.x^{\frac{3}{2}}-\frac{x^3}{36}{]}_3^{12}$
$=\frac{147}{4}$...(ii)
Hence
Ratio: $\frac{i}{ii}$
$=\frac{45}{147}$
$=\frac{15}{49}$.