Find the directrices of the hyperbola.
-$$81 x ^ { 2 } - 16 y ^ { 2 } = 1296$$
A) $\mathrm { y } = \pm \frac { 81 } { \sqrt { 97 } }$
B) $y = \pm \frac { 16 } { \sqrt { 97 } }$
C) $x = \pm \frac { 16 } { \sqrt { 97 } }$
D) $x = \pm \frac { 81 } { \sqrt { 97 } }$

Question

Quizplus · Accepted Answer

The equation of the hyperbola is given by $81x^2 - 16y^2 = 1296$. To find the directrices, we first need to bring the equation to its standard form. The standard form of a hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $a$ and $b$ are constants. Dividing the given equation by 1296, we get $\frac{x^2}{16} - \frac{y^2}{81} = 1$, which means $a^2 = 16$ and $b^2 = 81$. The directrices of a hyperbola are given by $x = \pm \frac{a^2}{e}$, where $e$ is the eccentricity of the hyperbola. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{81}{16}} = \sqrt{\frac{97}{16}}$. Therefore, the directrices are $x = \pm \frac{a^2}{e} = \pm \frac{16}{\sqrt{\frac{97}{16}}} = \pm \frac{16}{\sqrt{97}} \cdot \frac{4}{4} = \pm \frac{16}{\sqrt{97}}$.

Find the Directrices of the Hyperbola 81x2−16y2=129681 x ^ { 2 } - 16 y ^ { 2 } = 129681x2−16y2=1296

Find the Directrices of the Hyperbola $81 x ^ { 2 } - 16 y ^ { 2 } = 1296$