Question

The equation of the hyperbola in the standard form (with transverse axis along the x-axis) having the length of the latus rectum = 9 unit and eccentricity = $\frac{5}{4}$, is-

• A. $\frac{x^{2}}{16} - \frac{y^{2}}{18} = 1$
• B. $\frac{x^{2}}{36} - \frac{y^{2}}{27} = 1$
• C. $\frac{x^{2}}{64} - \frac{y^{2}}{36} = 1$
• D. $\frac{x^{2}}{36} - \frac{y^{2}}{64} = 1$

Answer 1

Answer: (C)

$\frac{x^{2}}{64} - \frac{y^{2}}{36} = 1$

Answer 2

Since, the transverse axis is the x-axis, then the equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}$ = 1, (a > b).

Length of latus rectum = 9 = $\frac{2b^{2}}{a}$ … (i)

and e = $\frac{5}{4}$Ž 1 + $\frac{b^{2}}{a^{2}}$ = $\frac{25}{16}$

Ž 1 + $\frac{9a}{2a^{2}}$= $\frac{25}{16}$ [using Equation (i)]

Ž $\frac{9}{2a}$= $\frac{9}{16}$ Ž a = 8

On putting this value in Equation (i), we get

b² = $\frac{9 \times 8}{2}$ = 36 Ž b = 6

\ Equation of hyperbola is

$\frac{x^{2}}{8^{2}} - \frac{y^{2}}{6^{2}}$ = 1 Ž $\frac{x^{2}}{64}$ – $\frac{y^{2}}{36}$ = 1.