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_{P Q}=\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+a b$
$ y=(a-b) x+a b$
$\therefore \left.S_1=\int\limits_{-b}^a(a-b) x+a b-x^2\right) d x$
which simplifies to $\frac{(a+b)^3}{6}$ .....(1)
Also $S _2=\frac{1}{2}\begin{vmatrix}a & a ^2 & 1 \\ - b & b ^2 & 1 \\ 0 & 0 & 1\end{vmatrix}=\frac{1}{2}\left[ ab ^2+ a ^2 b \right]=\frac{1}{2} ab ( a + b )$....(2)
$\therefore \frac{ S _1}{ S _2}=\frac{( a + b )^3}{6} \cdot \frac{2}{ ab ( a + b )}=\frac{( a + b )^2}{3 ab }=\frac{1}{3}\left[\frac{ a }{ b }+\frac{ b }{ a }+2\right] $
$\left.\therefore \frac{ S _1}{ S _2}\right|_{\min .}=\frac{4}{3} $