JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If the tangents drawn to the hyperbola \(4y^2 = x^2 + 1\) intersect the co-ordinate axes at the distinct points \(A\) and \(B\), then the locus of the mid point of \(AB\) is
- A \(x^2 - 4y^2 + 16 x^2y^2 = 0\)
- B \(4x^2 -y^2 + 16 x^2 y^2 = 0\)
- C \(4x^2 -y^2 - 16 x^2 y^2 = 0\)
- D \(x^2 - 4y^2 - 16 x^2 y^2 = 0\)
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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