Question

Area of segment cut off from the parabola ${{x}^{2}}=8y$ by the line $x-2y+8=0$ is
$\begin{align}
  & \text{a) 12} \\
 & \text{b) 24} \\
 & \text{c) 48} \\
 & \text{d) 36} \\
\end{align}$

Answer 1

Hint: We know that the equation of the parabola is ${{x}^{2}}=8y$ and the line is $x-2y+8=0$
Hence we will solve the equations to get point of intersection.
Once we have point of intersection we will use the condition to find the required area by integrating the equation.

Complete step by step answer:
Now first consider the equation ${{x}^{2}}=8y.................(1)$
and the line $x-2y+8=0$
We will find the intersection point of the line and parabola.
Now we can also write the equation of line as $x+8=2y$ .
Dividing the equation by 2 we get
$y=\dfrac{x+8}{2}$
Now let us substitute the value of y from equation (1) and substitute it in equation (1)
Hence we get
$\begin{align}
  & {{x}^{2}}=8\left( \dfrac{x+8}{2} \right) \\
 & \Rightarrow {{x}^{2}}=4(x+8) \\
 & \Rightarrow {{x}^{2}}=4x+32 \\
 & \Rightarrow {{x}^{2}}-4x-32=0 \\
\end{align}$
Now we have a quadratic equation in x and hence we will solve it to find the values of x.
Let us write - 4x as 4x – 8x, hence we get
${{x}^{2}}+4x-8x-32=0$
$\Rightarrow x(x+4)-8(x+4)=0$
$\Rightarrow (x-8)(x+4)=0$
Now we know that if a.b = 0, then either a = 0 or b = 0.
Hence we get either (x – 8) = 0 or (x + 4) = 0.
Hence the value of x is 8 or – 4.
Now at x = - 4 we get $y=\dfrac{-4+8}{2}=2$
And similarly if x = 8 we have $y=\dfrac{8+8}{2}=8$
Hence the point of intersection are $(4,6)$ and $(8,8)$ .

Now we know the area under the curve can be found by integration.
To find the area cut by the line and the curve we will have to find the area under line and subtract it by area under curve all from point x = -4 to x = 8
Now equation of line is $y=\dfrac{x+8}{2}$ and equation of parabola is $y=\dfrac{{{x}^{2}}}{8}$
Hence the required area is \[\int\limits_{-4}^{8}{\left( \dfrac{x+8}{2}-\dfrac{{{x}^{2}}}{8} \right)}\]
\[=\int\limits_{-4}^{8}{\left( \dfrac{x+8}{2} \right)dx-\int\limits_{-4}^{8}{\left( \dfrac{{{x}^{2}}}{8} \right)dx}}\]
\[=\int\limits_{-4}^{8}{\left( \dfrac{x}{2} \right)dx}+\int\limits_{-4}^{8}{\left( \dfrac{8}{2} \right)dx}-\int\limits_{-4}^{8}{\left( \dfrac{{{x}^{2}}}{8} \right)dx}\]
\[\begin{align}
  & ={{\left[ \dfrac{{{x}^{2}}}{4} \right]}^{8}}_{-4}+\dfrac{8}{2}{{\left[ x \right]}^{8}}_{-4}-\dfrac{1}{8}{{\left[ \dfrac{{{x}^{3}}}{3} \right]}^{8}}_{-4} \\
 & =\dfrac{{{8}^{2}}}{4}-\dfrac{{{4}^{2}}}{4}+\dfrac{8(8)}{2}-\dfrac{8\left( -4 \right)}{2}-\dfrac{1}{8}\left( \dfrac{{{8}^{3}}}{3} \right)+\dfrac{1}{8}\left( \dfrac{-{{4}^{3}}}{3} \right) \\
 & =16-4+32+16-\dfrac{1}{8}\left( \dfrac{512}{3}+\dfrac{64}{3} \right) \\
 & =12+48-\dfrac{1}{8}\left( \dfrac{512+64}{3} \right) \\
\end{align}\]
$\begin{align}
  & =60-\left( \dfrac{576}{24} \right) \\
 & =60-24 \\
 & =36 \\
\end{align}$
Hence the area under the curve is 36 sq units
Option d is the correct option.
Note:
 Note that here the line is above the curve and hence while finding area we take integration of line – curve and not curve – line. Look at the diagram for such questions to be clear