For the parabola $y=-x^2$,let $a<0,b>0,$ $P(a,-a^2),Q(b,-b^2)$.Let $M$ be the mid point of $PQ$ and $R$ be the point of intersection of the vertical line through $M,$ with the parabola.Find the ratio of the area of the region bounded by the parabola and the line segment $PQ$ to the area of triangle $PQR$
By integration,i found the area of the region bounded by the parabola and the line segment $PQ=\frac{-1}{3}(a^3-b^3)$ and the point $R=(\frac{a+b}{2},-(\frac{a+b}{2})^2)$
I am stuck here.
If $H$ is the projection of $P$ on line $RM$ and $K$ is the projection of $Q$ on line $RM$, then the area of triangle $PQR$ is given by $$ A_{PQR}={1\over2}RM\cdot PH+{1\over2}RM\cdot QK ={1\over2}RM\cdot (b-a). $$ Hence you only need to compute $\displaystyle RM={a^2+b^2\over2}-\left({a+b\over2}\right)^2$.