JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let one focus of the hyperbola \(\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) be at \((\sqrt{10}, 0)\) and the corresponding directrix be \(\mathrm{x}=\frac{9}{\sqrt{10}}\). If e and \(l\) respectively are the eccentricity and the length of the latus rectum of H , then \(9\left(\mathrm{e}^2+l\right)\) is equal to:
- A 14
- B 15
- C 16
- D 12
Answer & Solution
Correct Answer
(C) 16
Step-by-step Solution
Detailed explanation
\(\mathrm{ae}=\sqrt{10} \text { and } \frac{\mathrm{a}}{\mathrm{e}}=\frac{9}{10} \) \( \Rightarrow \mathrm{a}^2=9 \text { and } \mathrm{e}=\frac{\sqrt{10}}{3} \) Now \((\mathrm{ae})^2=\mathrm{a}^2+\mathrm{b}^2 \) \( 10=9+\mathrm{b}^2 \Rightarrow \mathrm{~b}^2=1 \)…
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