Question

The area bounded by the curve $x^2=4y+4$ and line $3x+4y=0$ is

• A. $\frac{25}{4}$ sq. units
• B. $\frac{125}{8}$ sq. units
• C. $\frac{125}{16}$ sq. units
• D. $\frac{125}{24}$ sq. units

Answer 1

Answer: (D)

We have, $x^2=4y+4\dots \left(i\right)$
and $3x+4y=0\dots \left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get $x=-4,1$

Required area = area of shaded region.
$={\int }_{-4}^1\left(-\frac{3x}{4}-\frac{x^2}{4}+1\right)dx$
$=-\frac{3}{8}\left(1-16\right)-\frac{1}{12}\left(1+64\right)+5$
$=\frac{45}{8}-\frac{5}{12}$
$=\frac{125}{24}$ sq. units