Three points O(0,0), P(a, a^2), Q(-b, b^2), a\gt 0, ~b\gt 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 OPQ . If the minimum value of \fracS_1~S_2 is \fracmn, \operatornamegcd(m, n)=1 , then m+n is equal to \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q51 with Solution

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Correct answer: 7

Given: $O(0,0), \quad P(a,a^2), \quad Q(-b,b^2), \quad a>0,\; b>0$ all lie on the parabola $y=x^2.$

We need:

Then minimize $\frac{S_1}{S_2}.$

Slope of $PQ$ is $m=\frac{a^2-b^2}{a-(-b)}=\frac{(a-b)(a+b)}{a+b}=a-b.$

Using point $P(a,a^2)$ , equation of the line is $y-a^2=(a-b)(x-a).$ So, $y=(a-b)x+ab.$

Thus chord $PQ$ is $y=(a-b)x+ab.$

The parabola is $y=x^2$ and the chord is above it between $x=-b$ and $x=a$ . Hence $S_1=\int_{-b}^{a}\left[(a-b)x+ab-x^2\right]dx.$

Now factor the integrand: $(a-b)x+ab-x^2 = -(x^2-(a-b)x-ab).$ Since $x^2-(a-b)x-ab=(x-a)(x+b),$ we get $(a-b)x+ab-x^2 = -(x-a)(x+b)=(a-x)(x+b).$

So, $S_1=\int_{-b}^{a}(a-x)(x+b)\,dx.$

Expand: $(a-x)(x+b)=ax+ab-x^2-bx=(a-b)x+ab-x^2.$

Integrating directly, $S_1=\int_{-b}^{a}\left((a-b)x+ab-x^2\right)dx.$

Antiderivative: $\frac{(a-b)x^2}{2}+abx-\frac{x^3}{3}.$

Evaluate from $-b$ to $a$ :

At $x=a$ :

=\frac{a^3-a^2b}{2}+a^2b-\frac{a^3}{3} =\frac{a^3}{6}+\frac{a^2b}{2}.$$ At $x=-b$: $$\frac{(a-b)b^2}{2}-ab^2+\frac{b^3}{3} =\frac{ab^2-b^3}{2}-ab^2+\frac{b^3}{3} =-\frac{ab^2}{2}-\frac{b^3}{6}.$$ Therefore, $$S_1=\left(\frac{a^3}{6}+\frac{a^2b}{2}\right)-\left(-\frac{ab^2}{2}-\frac{b^3}{6}\right) =\frac{a^3+3a^2b+3ab^2+b^3}{6}.$$ Hence $$S_1=\frac{(a+b)^3}{6}.$$ --- 4. **Area $S_2$ of triangle $OPQ$** Using determinant formula for area of triangle with one vertex at origin: $$S_2=\frac12\left|x_Py_Q-x_Qy_P\right|.$$ Here $$P(a,a^2),\quad Q(-b,b^2).$$ So, $$S_2=\frac12\left|a\cdot b^2-(-b)\cdot a^2\right| =\frac12\left(ab^2+a^2b\right).$$ Thus, $$S_2=\frac12 ab(a+b).$$ --- 5. **Form the ratio** $$\frac{S_1}{S_2}=\frac{\frac{(a+b)^3}{6}}{\frac12 ab(a+b)} =\frac{(a+b)^2}{3ab}.$$ So we need to minimize $$R=\frac{(a+b)^2}{3ab}.$$ Rewrite: $$R=\frac{a^2+2ab+b^2}{3ab}=\frac13\left(\frac{a}{b}+2+\frac{b}{a}\right).$$ Let $$t=\frac{a}{b}>0.$$ Then $$R=\frac13\left(t+2+\frac1t\right).$$ Using AM-GM, $$t+\frac1t\ge 2,$$ with equality when $t=1$, i.e. $a=b$. Therefore, $$R_{\min}=\frac13(2+2)=\frac43.$$ Thus, $$\frac{m}{n}=\frac43,$$ so $$m=4,\quad n=3.$$ Hence, $$m+n=7.$$ --- 6. **Comparison with stored answer** Derived answer: $7$ Stored correct answer: $7$ They agree.