JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
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 \(4\)
- C \(7\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(7\)
Step-by-step Solution
Detailed explanation
\(\left.\mathrm{S}_2=1 / 2\left|\begin{array}{ccc}0 & 0 & 1 \\ \mathrm{a} & \mathrm{a}^2 & 1 \\ -\mathrm{b} & \mathrm{b}^2 & 1\end{array}\right | \right \rvert \,=1 / 2\left(a \mathrm{~b}^2+\mathrm{a}^2 \mathrm{~b}\right)\) \( P Q:-y-a^2=\frac{a^2-b^2}{a+b}(x-a) \)…
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