MCQ
Three points $\mathrm{O}(0,0), \mathrm{P}\left(\mathrm{a}, \mathrm{a}^2\right), \mathrm{Q}\left(-\mathrm{b}, \mathrm{b}^2\right), \mathrm{a}>0, \mathrm{~b}>0$, are on the parabola $y=x^2$. Let $S_1$ be the area of the region bounded by the line $P Q$ and the parabola, and $S_2$ be the area of the triangle $O P Q$. If the minimum value of $\frac{\mathrm{S}_1}{\mathrm{~S}_2}$ is $\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to :
  • A
    $65$
  • B
    $4$
  • $7$
  • D
    $6$

Answer

Correct option: C.
$7$
c
$\left.\mathrm{S}_2=1 / 2\left|\begin{array}{ccc}0 & 0 & 1 \\ \mathrm{a} & \mathrm{a}^2 & 1 \\ -\mathrm{b} & \mathrm{b}^2 & 1\end{array}\right | \right \rvert \,=1 / 2\left(a \mathrm{~b}^2+\mathrm{a}^2 \mathrm{~b}\right)$

$ P Q:-y-a^2=\frac{a^2-b^2}{a+b}(x-a) $

$ y-a^2=(a-b) x-(a-b) a $

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

$ S_1=\int_{-b}^a\left((a-b) x+a b-x^2\right) d x $

$ =(a-b) \frac{x^2}{2}+(a b) x-\left.\frac{x^3}{3}\right|_{-b} ^a $

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

$ \frac{\mathrm{S}_1}{\mathrm{~S}_2}=\frac{\frac{(\mathrm{a}-\mathrm{b})^2}{2}+\mathrm{ab}-\frac{\left(\mathrm{a}^2+\mathrm{b}^2-\mathrm{ab}\right)}{3}}{\frac{\mathrm{ab}}{2}} $

$ =\frac{3(a-b)^2+6 a b-2\left(a^2+b^2-a b\right)}{3 a b} $

$ =\frac{1}{3}\left[\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}+2\right] $

$ =\frac{4}{3}=\frac{\mathrm{m}}{\mathrm{n}} \quad \mathrm{m}+\mathrm{n}=7 $

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