Question

The area of the tangent cut off from the parabola x2=8yx2=8y is:

• A. $32$
• B. $34$
• C. $36$
• D. $38$

Answer 1

The correct option is C $36$

Given parabola is $x^2=8y$ .....................$\left(1\right)$
and the straight line is $x-2y+8=0$ ......................$\left(2\right)$
Substituting the value of $y$ from $\left(2\right)$ in $\left(1\right)$ we get
$x^2=4(x+8)$ or $x^2-4x-32=0$
or $(x-8)(x+4)=0$
$\therefore x=8,-4$
Thus $\left(1\right)$ and $\left(2\right)$ intersect at $P$ and $Q$ where $x=8$ and $x=-4$
$\therefore$ Required area $POQ$ (i.e., dotted area)$=$ area bounded by straight line $\left(2\right)$ and $x-$axis from $x=-4$ to $x=8-$area bounded by parabola $\left(1\right)$ and $x-$axis from $x=-4$ to $x=8$
$={\int }_{-4}^{8}y.dx,$ from $\left(2\right)-{\int }_{-4}^{8}y.dx$, from $\left(1\right)$
$={\int }_{-4}^8\left(\frac{x+8}{2}\right)dx-{\int }_{-4}^8\frac{x^2}{8}dx$
$={[\frac{x^2}{4}+4x]}_{-4}^8-\frac{1}{8}{\left[\frac{x^3}{3}\right]}_{-4}^8$
$=\frac{1}{2}[(32+64)-(-24)]-\frac{1}{24}(512+64)$
$=36$