JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The tangent at an extremity (in the first quadrant) of latus rectum of the hyperbola \(\frac{{{x^2}}}{4} - \frac{{{y^2}}}{5} = 1\) , meet \(x-\) axis and \(y-\) axis at \(A\) and \(B\) respectively. Then \((OA)^2 - (OB)^2\) , where \(O\) is the origin, equals
- A \( - \frac{{20}}{9}\)
- B \( \frac{{16}}{9}\)
- C \(4\)
- D \( - \frac{{4}}{3}\)
Answer & Solution
Correct Answer
(A) \( - \frac{{20}}{9}\)
Step-by-step Solution
Detailed explanation
Given \(\frac{{{x^2}}}{4} - \frac{{{y^2}}}{5} = 1\) \( \Rightarrow {a^2} = 4,{b^2} = 5\) \(e = \sqrt {\frac{{{a^2} + {b^2}}}{{{a^2}}}} = \sqrt {\frac{{4 + 5}}{4}} = \frac{3}{2}\) \(L = \left( {2 \times \frac{3}{2},\frac{5}{2}} \right) = \left( {3,\frac{5}{2}} \right)\) equation…
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