Question

Find the area of the region bounded by $y^2=2x$ and the line $x-y=4$

Answer 1

The parabola $y^2=2x$ opens towards the positive $x-$axis and its focus is $(\frac{1}{2},0)$

The straight line $x-y=4$ passes through $(4,0)$ and $(0,-4)$

Solving $y^2=2x$ and $x-y=4$, we get

$y^2=2(y+4)$

$\Rightarrow y^2-2y-8=0$

$\Rightarrow (y-4)(y+2)=0$

$\Rightarrow y=4$ or $y=-2$

So, the points of intersection of the given parabola and the line are $A(8,4)$ and $B(2,-2)$

$\therefore$ Required area$=$Area of the shaded region $OABO$

$={\int }_{-2}^4{x}_{line}dy-{\int }_{-2}^4{x}_{parabola}dy$

$={\int }_{-2}^4(y+4)dy-{\int }_{-2}^4\frac{y^2}{2}dy$

$={\left[\frac{{(y+4)}^2}{2}\right]}_{-2}^4-\frac{1}{2}{\left[\frac{y^3}{3}\right]}_{-2}^4$

$=\frac{1}{2}(64-4)-\frac{1}{6}(64-(-8))$

$=30-12=18$sq.units.