JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The normal to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1\) at the point \((8,3 \sqrt{3})\) on it passes through the point
- A \((15,-2 \sqrt{3})\)
- B \((9,2 \sqrt{3})\)
- C \((-1,9 \sqrt{3})\)
- D \((-1,6 \sqrt{3})\)
Answer & Solution
Correct Answer
(C) \((-1,9 \sqrt{3})\)
Step-by-step Solution
Detailed explanation
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1:(8,3 \sqrt{3})\) lie on Hyperbola then \(\frac{64}{a^{2}}-\frac{27}{9}=1 \Rightarrow a^{2}=\frac{64}{4}=16\) equation of normal at \((8,3 \sqrt{3})\) : \(\frac{16 x}{8}+\frac{9 y}{3 \sqrt{3}}=16+9\) \(2 x+\sqrt{3} y=25\)
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