Question

If the area bounded by the curve $4x^2-8x+y^2-2y+1=0$ is divided into two parts by straight line $2x+y=5,$ then the ratio of smaller to the larger area is

• A. $\left(\pi -2\right):\left(3\pi +2\right)$
• B. $\left(\pi +2\right):\left(2\pi +1\right)$
• C. $\left(3\pi -2\right):\left(3\pi +2\right)$
• D. $\left(2\pi +2\right):\left(5\pi -2\right)$

Answer 1

The correct option is A $\left(\pi -2\right):\left(3\pi +2\right)$

Given curve: $4x^2-8x+y^2-2y+1=0$ can be converted into
$\Rightarrow \frac{(x-1{)}^2}{1}+\frac{(y-1{)}^2}{4}=1$
Shifting of the curve does not change the area.

Here, origin is shifted at $O_1(1,1)$
So area of ellipse $=\pi (1\times 2)=2\pi$
and the line $2x+y=5$ can be written as,
$\frac{x-1}{1}+\frac{y-1}{2}=1$
$\Rightarrow$Smaller area $=\frac{2\pi }{4}-$ (area of triangle formed in $1^{st}$ quadrant)
$=\frac{\pi }{2}-\frac{1}{2}\cdot 1\cdot 2=\frac{\pi -2}{2}$
and larger area $=2\pi -\frac{\pi -2}{2}=\frac{3\pi +2}{2}$
So, ratio $=\frac{\pi -2}{3\pi +2}$