Question

Find the ratio in which the area bounded by the curves y2 = 12x and x2 = 12y is divided by the line x = 3.

Answer 1

Solution:

The Three curves are , $y^2= 12 x, x^2= 12 y$ and x=3.

Point of intersection of , $y^2=1 2 x  \text{and} , x=3 is$

y²= 12 × 3

y²= 36

y= $\pm6$

Similarly, point of intersection of , $x^2=1 2 y  \text{and} , x=3 is$

9 = 12 y

$y=\frac{9}{12}\\\\y=\frac{3}{4}$

Area under the curve ,  $y^2=1 2 x  \text{and} , x=3 is$

$=\int\limits^3_0 {\sqrt{12x}} \, dx \\\\ =\sqrt{12}\int\limits^3_0 {\sqrt{x}} \, dx\\\\ =2\sqrt{3}\frac{x^\frac{3}{2}}{\frac{3}{2}}\left \{ {{x=3} \atop {x=0}} \right. \\\\ = \frac{4}{\sqrt3}\times 3^{\frac{3}{2}}\\\\ = 4 \times 3 \\\\ = 12$

Area under the curve , $x^2=1 2 y  \text{and} , x=3 is$

$=\int\limits^3_0 {\frac{x^2}{12}} \, dx \\\\ = \frac{1}{12 \times 3}[x^3]\left \{ {{x=3} \atop {x=0}} \right. \\\\ =  \frac{1}{36}\times 3^3\\\\ =\frac{3}{4}$

The ratio in which the area bounded by the curves y² = 12 x and x² = 12 y is divided by the line x = 3 is

  $=\frac{2 \times 12}{\frac{3}{4}}\\\\ =\frac{96}{3}\\\\ =32:1$