Find the eccentricity e of the hyperbola.
-$16 x ^ { 2 } - y ^ { 2 } = 1$
A) 4
B) $\frac { \sqrt { 17 } } { 4 }$
C) $\sqrt { 17 }$
D) 17

Question

Quizplus · Accepted Answer

The standard form of a hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Comparing this with the given equation $16x^2 - y^2 = 1$, we can rewrite it as $\frac{x^2}{(1/16)} - \frac{y^2}{1} = 1$, which implies $a^2 = \frac{1}{16}$ and $b^2 = 1$. The eccentricity $e$ of a hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$. Substituting $a^2 = \frac{1}{16}$ and $b^2 = 1$, we get $e = \sqrt{1 + 16} = \sqrt{17}$.

Find the Eccentricity E of the Hyperbola 16x2−y2=116 x ^ { 2 } - y ^ { 2 } = 116x2−y2=1

Find the Eccentricity E of the Hyperbola $16 x ^ { 2 } - y ^ { 2 } = 1$