Mathematics - Area Under Curves Question with Solution | TestHub

Let $C(h,h^2)$ and $D(k,k^2)$.

Because $M_{CD}=2\Rightarrow k+h=2$.

Also, $CD=DA$.

$\begin{array}{c}\sqrt{5}|k-h|=\frac{h^2-2h+17}{\sqrt{5}},\\ k=2-h.\\ t=|h-1|:t^2-10t+16=0,\\ t=2\text{ or }8.\\ C=(-1,1),D=(3,9)\end{array}$

or $(-7,49)$ and $D(9,81)$.

(a) The equation for CD is $(y-h^2)=2(x-h)$, so the y-intercept is $h^2-2h$.

The minimum value is 3 (when $h=-1$).

(b) The area of ABCD is $C{D}^2=5(k-h{)}^2$. The maximum area is $5(16{)}^2=1280$.

(c) For CD, $y=2x+3$. The area is ${\int }_{-1}^3(2x+3-x^2)dx=\frac{32}{3}$.