Question

3 points $O(0,0),P\left(a,a^2\right),Q\left(-b,b^2\right)(a>0,b>0)$ are on the parabola $y=x^2$. Let $S_1$ be the area bounded by the line $PQ$ and the parabola and let $S_2$ be the area of the triangle $OPQ$, the minimum value of $S_1/S_2$ is

• A. 4 / 3
• B. $5/3$
• C. 2
• D. $7/3$

Answer 1

Answer: (A)

$m_{PQ}=\frac{a^2-b^2}{a+b}=a-b$
equation of $PQ$

$y-a^2=\frac{a^2-b^2}{a+b}(x-a)$
$\text{ or }y-a^2=(a-b)(x-a)$
$y=a^2+x(a-b)-a^2+ab$
$y=(a-b)x+ab$
$\therefore S_1=\underset{-b}{\overset{a}{\int }}(a-b)x+ab-x^2)dx$
which simplifies to $\frac{(a+b{)}^3}{6}$ .....(1)
Also $S_2=\frac{1}{2}\left|\begin{array}{ccc}a& a^2& 1\\ -b& b^2& 1\\ 0& 0& 1\end{array}\right|=\frac{1}{2}\left[a{b}^2+a^{2}b\right]=\frac{1}{2}ab(a+b)$....(2)
$\therefore \frac{S_1}{S_2}=\frac{(a+b{)}^3}{6}\cdot \frac{2}{ab(a+b)}=\frac{(a+b{)}^2}{3ab}=\frac{1}{3}\left[\frac{a}{b}+\frac{b}{a}+2\right]$
${\therefore \frac{S_1}{S_2}|}_{\min .}=\frac{4}{3}$