Comment
Share
Q)

# Equation of the hyperbola with eccentricity $\large\frac{3}{2}$ and foci at $(\pm 2,0)$ is

$\begin{array}{1 1}\large\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}\\\large\frac{x^2}{9}-\frac{y^2}{9}=\frac{4}{9}\\\large\frac{x^2}{4}-\frac{y^2}{9}\normalsize =1\\\large\frac{x^2}{2}-\frac{y^2}{3}\normalsize =1\end{array}$

Comment
A)
Toolbox:
• General equation of hyperbola which lies on the $x$-axis is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1 • b^2=a^2(e^2-1) Answer : \large\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9} Given eccentricity 'e'=\large\frac{3}{2} Coordinate of foci =(\pm 2,0) It is clear that the hyperbola lies on the x-axis. Hence its equation is \large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$
$ae=2$
$\therefore a(\large\frac{3}{2})$$=2 a=\large\frac{4}{3} a^2=\large\frac{16}{9} b^2=a^2(e^2-1) b^2=\large\frac{16}{9}(\frac{9}{4}$$-1)$
$\;\;\;\;=\large\frac{16}{9}\times \frac{5}{4}$
$\;\;\;\;=\large\frac{20}{9}$
Hence the equation of the hyperbola is $\large\frac{9x^2}{16}-\frac{9y^2}{20}$$=1$
(ie) $\large\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$