JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If a directrix of a hyperbola centered at the origin and passing through the point \((4, -2\sqrt 3)\) is \(5x = 4\sqrt 5\) and its eccentricity is \(e\), then
- A \(4e^4 + 8e^2 -35 = 0\)
- B \(4e^4 -24e^2 + 35 = 0\)
- C \(4e^4 -12e^2 -27 = 0\)
- D \(4e^4 -24e^2 + 27 = 0\)
Answer & Solution
Correct Answer
(B) \(4e^4 -24e^2 + 35 = 0\)
Step-by-step Solution
Detailed explanation
Let hyperbola be \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) and passes through \(\left( {4, - 2\sqrt 3 } \right)\) therefore \(\frac{{16}}{{{a^2}}} - \frac{{12}}{{{b^2}}} = 1\,\,\,\,\,\,\,\,\,\,\,......\left( i \right)\) \(\because\)…
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