JEE Main 2024 — 1 February, Shift 2 — Question 27

$$S_{2} =\frac12 \begin{vmatrix} 0 & 0 & 1\\[4pt] a & a^{2} & 1\\[4pt] -b & b^{2} & 1 \end{vmatrix}$$ $$=\frac12\Bigl( a b^{\,2} + a^{2} b \Bigr).$$

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

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

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

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

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

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

$\frac{\mathrm{S}_{1}}{\mathrm{~S}_{2}}=\frac{\frac{(\mathrm{a}-\mathrm{b})^{2}}{2}+\mathrm{ab}-\frac{\left(\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{ab}\right)}{3}}{\frac{\mathrm{ab}}{2}}$

$=\frac{3(\mathrm{a}-\mathrm{b})^{2}+6 \mathrm{ab}-2\left(\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{ab}\right)}{3 \mathrm{ab}}$ $=\frac{1}{3}\left[\frac{\mathrm{a}}{\underset{\operatorname{min}=2}{\mathrm{b}}}+\frac{\mathrm{b}}{\mathrm{a}}+2\right]$ $=\frac{4}{3}=\frac{\mathrm{m}}{\mathrm{n}}$

$\quad \mathrm{m}+\mathrm{n}=7$