JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the eccentricity of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) be \(\frac{5}{4}\). If the equation of the normal at the point \(\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)\) on the hyperbola is \(8 \sqrt{5} x +\beta y =\lambda\), then \(\lambda-\beta\) is equal to
- A \(89\)
- B \(85\)
- C \(78\)
- D \(45\)
Answer & Solution
Correct Answer
(B) \(85\)
Step-by-step Solution
Detailed explanation
\(e ^{2}=1+\frac{ b ^{2}}{ a ^{2}}=\frac{25}{16} \Rightarrow \frac{ b ^{2}}{ a ^{2}}=\frac{9}{16} \ldots \ldots(1)\) \(A \left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)\) satisfies \(\frac{ x ^{2}}{ a ^{2}}-\frac{ y ^{2}}{ b ^{2}}=1\)…
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