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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

माना कि दीर्घवृत्त \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \) \( (a>b) \) के नाभिलंब की लंबाई 30 है। यदि इसकी उत्केंद्रता फलन \( f(t)=-\frac{3}{4}+2t-t^{2} \) का अधिकतम मान है, तो \( (a^{2}+b^{2}) \) = ___ है।

  1. A 516
  2. B 256
  3. C 496
  4. D 276
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Answer & Solution

Correct Answer

(C) 496

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Detailed explanation

\( f(t)=\frac{-3}{4}+2t-t^{2} \) \( f(t)|_{maximum}=\frac{1}{4}=e \Rightarrow e^{2}=\frac{1}{16} \Rightarrow \frac{a^{2}-b^{2}}{a^{2}}=\frac{1}{16} \) \(\quad\ldots(1)\) \(\because \frac{2b^{2}}{a}=30 \Rightarrow b^{2}=15a \quad\ldots(2)\) By (1) & (2)…
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