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

एक अविपरवलय बिंदु \(P(\sqrt{2}, \sqrt{3})\) से होकर जाता है, तथा उसकी नाभियाँ \((\pm 2,0)\) पर है, तो अतिपरवलय के बिंदु \(P\) पर खींची गई स्पर्शरिखा जिस बिंदु से होकर जाती है, वह है:

  1. A \(\left( { - \sqrt 2 , - \sqrt 3 } \right)\)
  2. B \(\left( {3\sqrt 2 ,2\sqrt 3 } \right)\)
  3. C \(\left( {2\sqrt 2 ,3\sqrt 3 } \right)\)
  4. D \(\left( {3,\sqrt 2 } \right)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left( {2\sqrt 2 ,3\sqrt 3 } \right)\)

Step-by-step Solution

Detailed explanation

Equation of hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) foci is \((\pm 2,0) \Rightarrow a c=2 \Rightarrow a^{2} c^{2}=4\) Since \(b^{2}=a^{2}\left(e^{2}-1\right)\) \(b^{2}=a^{2} e^{2}-a^{2}\) \(\therefore \quad a^{2}+b^{2}=4\) .......\((1)\) Hyperbola passes…
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