Find the equation of the hyperbola

Question:

Find the equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0).

Solution:

Given

$e=\frac{3}{2}$

We have foci $=(\pm \mathrm{a} \mathrm{e}, 0)=(\pm 2,0)$

Therefore the hyperbola lies on $x$-axis,

Equation is $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$

Given $\mathrm{a} \mathrm{e}=2$

$a \times \frac{3}{2}=2$

$a=\frac{4}{3}$

$\because b^{2}=a^{2}\left(e^{2}-1\right)$

$b^{2}=\left(\frac{4}{3}\right)^{2}\left(\left(\frac{3}{2}\right)^{2}-1\right)$

$=\frac{16}{9}\left(\frac{9}{4}-1\right)=\frac{16}{9} \times \frac{5}{4}=\frac{20}{9}$

Equation is $\frac{x^{2}}{\left(\frac{4}{3}\right)^{2}}-\frac{y^{2}}{\frac{20}{9}}=1$

$\frac{9 x^{2}}{16}-\frac{9 y^{2}}{20}=1$

Hence, the required equation is $\frac{9 \mathrm{x}^{2}}{16}-\frac{9 \mathrm{y}^{2}}{20}=1$

Equation is $\frac{\mathrm{x}^{2}}{\left(\frac{4}{3}\right)^{2}}-\frac{\mathrm{y}^{2}}{\frac{20}{9}}=1$

$\frac{9 x^{2}}{16}-\frac{9 y^{2}}{20}=1$

Hence, the required equation is $\frac{9 \mathrm{x}^{2}}{16}-\frac{9 \mathrm{y}^{2}}{20}=1$

Equation is $\frac{\mathrm{x}^{2}}{\left(\frac{4}{3}\right)^{2}}-\frac{\mathrm{y}^{2}}{\frac{20}{9}}=1$

$\frac{9 x^{2}}{16}-\frac{9 y^{2}}{20}=1$

Hence, the required equation is $\frac{9 \mathrm{x}^{2}}{16}-\frac{9 \mathrm{y}^{2}}{20}=1$