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

एक अतिपरवलय का केन्द्र मूलबिन्दु पर है तथा यह बिन्दु \((4,-2 \sqrt{3})\) से होकर जाता है। यदि इसकी एक नियता (directrix) \(5 x =4 \sqrt{5}\) है तथा इसकी उत्केन्द्रता \(e\) है, तो

  1. A \(4e^4 + 8e^2 -35 = 0\)
  2. B \(4e^4 -24e^2 + 35 = 0\)
  3. C \(4e^4 -12e^2 -27 = 0\)
  4. D \(4e^4 -24e^2 + 27 = 0\)
Verified Solution

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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