JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The vertices of a hyperbola \(H\) are \((\pm 6,0)\) and its eccentricity is \(\frac{\sqrt{5}}{2}\). Let \(N\) be the normal to \(H\) at a point in the first quadrant and parallel to the line \(\sqrt{2} x + y =2 \sqrt{2}\). If \(d\) is the length of the line segment of \(N\) between \(H\) and the \(y\)-axis then \(d ^2\) is equal to \(............\).
- A \(215\)
- B \(216\)
- C \(217\)
- D \(218\)
Answer & Solution
Correct Answer
(B) \(216\)
Step-by-step Solution
Detailed explanation
\(H : \frac{ x ^2}{36}-\frac{y^2}{9}=1\) equation of normal is \(6 x \cos \theta+3 y \cot \theta=45\) \(\text { slope }=-2 \sin \theta=-\sqrt{2}\) \(\Rightarrow \theta=\frac{\pi}{4}\) Equation of normal is \(\sqrt{2} x+y=15\) \(P:(a \sec \theta, b \tan \theta)\)…
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