Question:
A hyperbola has its centre at the origin, passes through the point $(4,2)$ and has transverse axis of length 4 along the $x$-axis. Then the eccentricity of the hyperbola is :
Correct Option: , 4
Solution:
Consider equation of hyperbola
$\frac{x^{2}}{2^{2}}-\frac{y^{2}}{b^{2}}=1$
$\because(4,2)$ lies on hyperbola
$\therefore \frac{16}{4}-\frac{4}{b^{2}}=1$
$\therefore b^{2}=\frac{4}{3}$Since, eccentricity $=\sqrt{1+\frac{b^{2}}{a^{2}}}$
Hence, eccentricity $=\sqrt{1+\frac{\frac{4}{3}}{4}}=\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.