Question

Find the area bounded by the curve $x^2=4y$ and the line $x=4y-2$. If the answer is $\frac{a}{8}$. what is the value of $a$?

Answer 1

Here,

$x^2=4y$ is a parabola.

And $x=4y-2$ is a line which intersects the parabola at points $A$ and $B$

We need to find area of shaded region.

First we are going to find points $A$ and $B.$

We know that,

$x=4y-2$

Putting in equation of curve, we get

$x^2=4y$

$(4y-2{)}^2=4y$

$16y^2+4-16y=4y$

$16y^2-16y-4y+4=0$

$16y^2-20y+4=0$

$4[4y^2-5y+1]=0$

$4y^2-5y+1=0$

$4y^2-4y-y+1=0$

$4y(y-1)-1(y-1)=0$

$(4y-1)(y-1)=0$

So, $y=\frac{1}{4},y=1$

For, $y=\frac{1}{4}:$

$x=4y-2$

$x=4\left(\frac{1}{4}\right)-2$

$x=-1$

So, point is $(-1,\frac{1}{4})$

For $y=1:$

$x=4y-2$

$x=4\left(1\right)-2$

$x=2$

So, point is $(2,1)$

As point $A$ is ine $2^{nd}$ quadrant

$\therefore$ $A=(-1,\frac{1}{4})$

And point $B$ is in $1^{st}$ quadrant

$\therefore$ $B=(2,1)$

Now,

$x=4y-2$

$y=\frac{x+2}{4}$

$\Rightarrow$ Area of $APBQ={\int }_{-1}^2\frac{x+2}{4}dx$

$=\frac{1}{4}{\int }_{-1}^2(x+2)dx$

$=\frac{1}{4}{[\frac{x^2}{2}+2x]}_{-1}^2$

$=\frac{1}{4}[(\frac{2^2}{2}+2(2\left)\right)-(\frac{(-1{)}^2}{2}+2(-1\left)\right)]$

$=\frac{1}{2}[6-\frac{1}{2}+2]$

$=\frac{1}{4}\times \frac{15}{2}$

$=\frac{15}{8}$

Now,

$x^2=4y$

So, $y=\frac{1}{4}{x}^2$

Area of $APOQBA=\frac{1}{4}{\int }_{-1}^2{x}^{2}dx$

$=\frac{1}{4}{\left[\frac{x^3}{3}\right]}_{-1}^2$

$=\frac{1}{4}\left[\frac{8+1}{3}\right]$

$=\frac{3}{4}$

Required area $=$ Area $APBQ-$ Area $APOQBA$

$=\frac{15}{8}-\frac{3}{4}$

$=\frac{15-6}{8}$

$=\frac{9}{8}$