Three points O ( 0 , 0 ) , P ( a , a 2 ) , Q ( - b , b 2 ) , a > 0 , b > 0 , are on the parabola y = x 2 . Let S 1 be the area of the region bounded by the line PQ and the parabola, and S 2 be the area of the triangle O P Q . If the minimum valu

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Solution

Q.
Three points $O (0, 0), P (a, a^{2}), Q (- b, b^{2}), a > 0, b > 0,$ are on the parabola $y = x^{2} .$ Let $S_{1}$ be the area of the region bounded by the line PQ and the parabola, and $S_{2}$ be the area of the triangle $O P Q .$ If the minimum value of $\frac{S_{1}}{S_{2}}$ is $\frac{m}{n}, g c d (m, n) = 1,$ then $m + n$ is equal to _______ . [2024]

Ans.
(7)

Equation of line $P Q$ is, $y - a^{2} = \frac{b^{2} - a^{2}}{- b - a} (x - a)$

$\Rightarrow y = x (a - b) + a b$

$∴ S_{1} = \int_{- b}^{a} (x (a - b) + a b - x^{2}) d x$

$= {[(a - b) \frac{x^{2}}{2} + a b x - \frac{x^{3}}{3}]}_{- b}^{a}$

$= (a - b) \frac{(a^{2} - b^{2})}{2} + a b (a + b) - \frac{(a^{3} + b^{3})}{3}$

$\Rightarrow S_{1} = \frac{1}{6} {(a + b)}^{3}$ ...(i)

Also, $S_{2} = \frac{1}{2}$ $| \begin{matrix} 0 & 0 & 1 \\ a & a^{2} & 1 \\ - b & b^{2} & 1 \end{matrix} |$ $= \frac{1}{2} a b (a + b)$

$Now, \frac{S_{1}}{S_{2}} = \frac{{(a + b)}^{3} / 6}{a b (a + b) / 2} = \frac{1}{3} \frac{{(a + b)}^{2}}{a b}$

$= \frac{a^{2} + b^{2} + 2 a b}{3 a b} = \frac{a}{3 b} + \frac{b}{3 a} + \frac{2}{3} = \frac{1}{3} (\frac{a}{b} + \frac{b}{a} + 2)$

$Now, \frac{a}{b} + \frac{1}{a / b} \geq 2$

$∴ Minimum value of \frac{S_{1}}{S_{2}} = \frac{4}{3}$

$So, m + n = 4 + 3 = 7$

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