JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The foci of a hyperbola are \(( \pm 2,0)\) and its eccentricity is \(\frac{3}{2}\). A tangent, perpendicular to the line \(2 x+3 y=6\), is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the \(x\) - and \(y\)-axes are \(a\) and \(b\) respectively, then \(|6 a|+|5 b|\) is equal to \(..........\).
- A \(11\)
- B \(12\)
- C \(13\)
- D \(10\)
Answer & Solution
Correct Answer
(B) \(12\)
Step-by-step Solution
Detailed explanation
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) \(ae =2\) and \(e =\frac{3}{2} \Rightarrow a =\frac{4}{3}\) also \(b ^2= a ^2 e ^2- a ^2 \Rightarrow 4-\frac{16}{9}\) \(\Rightarrow b^2=\frac{20}{9}\) \(\text { Slope of tangent }=\frac{3}{2}\) So tangent equation will be…
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