If length of transverse axis is 8 and conjugate axis is 10 and transverse axis is along x-axis then find the equation of hyperbola.(a) \((\frac{x}{4})^2-(\frac{y}{5})^2\)=1(b) \((\frac{x}{5})^2-(\frac{y}{4})^2\)=1(c) \((\frac{x}{10})^2-(\frac{y}{8})^2\)=1(d) \((\frac{x}{8})^2-(\frac{y}{10})^2\)=1I got this question in exam.Query is from Conic Sections topic in division Conic Sections of Mathematics – Class 11

If length of transverse axis is 8 and conjugate axis is 10 and transve

1.	If length of transverse axis is 8 and conjugate axis is 10 and transverse axis is along x-axis then find the equation of hyperbola.(a) \((\frac{x}{4})^2-(\frac{y}{5})^2\)=1(b) \((\frac{x}{5})^2-(\frac{y}{4})^2\)=1(c) \((\frac{x}{10})^2-(\frac{y}{8})^2\)=1(d) \((\frac{x}{8})^2-(\frac{y}{10})^2\)=1I got this question in exam.Query is from Conic Sections topic in division Conic Sections of Mathematics – Class 11
Answer» Right ANSWER is (a) \((\frac{x}{4})^2-(\frac{y}{5})^2\)=1 Easiest explanation: Given, 2a=8 => a=4 and 2b=10 => b=5. Equation of hyperbola with transverse AXIS ALONG x-axis is \((\frac{x}{4})^2-(\frac{y}{5})^2\)=1. So, equation of given hyperbola is \((\frac{x}{4})^2-(\frac{y}{5})^2\)=1.

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