JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the length of the latus rectum of an ellipse \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \) \( (a>b) \), be 30. If its eccentricity is the maximum value of the function \( f(t)=-\frac{3}{4}+2t-t^{2}, \) then \( (a^{2}+b^{2}) \) is equal to -
- A 516
- B 256
- C 496
- D 276
Answer & Solution
Correct Answer
(C) 496
Step-by-step Solution
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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