If foci of a hyperbola are (0, ±5) and length of semi transverse axis is 3 units, then find the equation of hyperbola.(a) \((\frac{x}{4})^2-(\frac{y}{3})^2\)=1(b) \((\frac{x}{3})^2-(\frac{y}{4})^2\)=1(c) \((\frac{x}{10})^2+(\frac{y}{8})^2\)=1(d) \((\frac{x}{8})^2-(\frac{y}{6})^2\)=1I got this question during an interview.This intriguing question comes from Conic Sections topic in section Conic Sections of Mathematics – Class 11

If foci of a hyperbola are (0, ±5) and length of semi transverse axis

1.	If foci of a hyperbola are (0, ±5) and length of semi transverse axis is 3 units, then find the equation of hyperbola.(a) \((\frac{x}{4})^2-(\frac{y}{3})^2\)=1(b) \((\frac{x}{3})^2-(\frac{y}{4})^2\)=1(c) \((\frac{x}{10})^2+(\frac{y}{8})^2\)=1(d) \((\frac{x}{8})^2-(\frac{y}{6})^2\)=1I got this question during an interview.This intriguing question comes from Conic Sections topic in section Conic Sections of Mathematics – Class 11
Answer» Right ANSWER is (a) \((\frac{x}{4})^2-(\frac{y}{3})^2\)=1 Explanation: GIVEN, a=3 and c=5 => B^2=c^2-a^2 = 5^2-3^2=4^2 => b=4. Equation of hyperbola with TRANSVERSE axis along y-axis is \((\frac{y}{a})^2-(\frac{x}{b})^2\)=1. So, equation of given hyperbola is \((\frac{y}{3})^2-(\frac{x}{4})^2\)=1.

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