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

यदि अतिपरवलय \(4 y ^{2}= x ^{2}+1\) पर खींची गई स्पर्शरिखाएँ निर्देशांक अक्षों को भिन्न बिंदुओं \(A\) तथा \(B\) पर काटती हैं, तो \(AB\) के मध्य्यबिंदु का बिंदुपथ है

  1. A \(x^2 - 4y^2 + 16 x^2y^2 = 0\)
  2. B \(4x^2 -y^2 + 16 x^2 y^2 = 0\)
  3. C \(4x^2 -y^2 - 16 x^2 y^2 = 0\)
  4. D \(x^2 - 4y^2 - 16 x^2 y^2 = 0\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(x^2 - 4y^2 - 16 x^2 y^2 = 0\)

Step-by-step Solution

Detailed explanation

equation of hyperbola is : \(4{y^2} = {x^2} + 1 \Rightarrow - {x^2} + 4{y^2} = 1\) \( \Rightarrow - \frac{{{x^2}}}{{{1^2}}} + \frac{{{y^2}}}{{{{\left( {\frac{1}{2}} \right)}^2}}} = 1\) \(\therefore a = 1,b = \frac{1}{2}\) Now, tangent to the curve at point…
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