Question

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

(a) $\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$

(b) $\frac{x^2}{9}-\frac{y^2}{9}=\frac{4}{9}$

(c) $\frac{x^2}{4}-\frac{y^2}{9}=1$

(d) none of these

Answer 1

$$\begin{gathered} \mathrm{For}\mathrm{an}\mathrm{hyperbola}, \\ \mathrm{eccentricity}\mathrm{is}\mathrm{given}\mathrm{as}\frac{3}{2}\mathrm{and}\mathrm{foci}\mathrm{at}\left(\pm 2,0\right) \\ \mathrm{i}.\mathrm{e}.\mathrm{hyperbola}\mathrm{has}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \mathrm{i}.\mathrm{e}.\pm ae=\pm 2 \\ \mathrm{i}.\mathrm{e}.ae=2 \\ \mathrm{i}.\mathrm{e}.a\times \frac{3}{2}=2 \\ \mathrm{i}.\mathrm{e}.a=\frac{4}{3},\mathrm{i}.\mathrm{e}.a^2=\frac{16}{9} \\ \therefore \mathrm{Since}{b}^2=a^2\left(e^2-1\right) \\ \Rightarrow b^2=\frac{16}{9}\left(\frac{9}{4}-1\right) \\ =\frac{16}{9}\times \left(\frac{5}{4}\right) \\ {b}^2=\frac{20}{9} \\ \mathrm{Hence},\mathrm{equation}\mathrm{of}\mathrm{hyperbola}\mathrm{is}\frac{x^2}{{\displaystyle \frac{16}{9}}}-\frac{y^2}{{\displaystyle \frac{20}{9}}}=1 \\ \mathrm{i}.\mathrm{e}.\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}; \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{option}\mathrm{A}. \end{gathered}$$