JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be \((-5,0)\) and \(5 x+9=0\), respectively. If the product of the focal distances of a point \((\alpha, 2 \sqrt{5})\) on the hyperbola is \(p\), then \(4 p\) is equal to
- A 180
- B 189
- C 190
- D 152
Answer & Solution
Correct Answer
(B) 189
Step-by-step Solution
Detailed explanation
Equation of hyperbola is \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) Directrix: \(\mathrm{x}=\frac{-9}{5}\) and corresponding foci \((-5,0)\) \(\Rightarrow-\frac{\mathrm{a}}{\mathrm{e}}=-\frac{9}{5}\) and \(-\mathrm{ae}=-5\)…
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