Find the standard-form equation of the hyperbola centered at the origin which satisfies the given conditions.
-Foci at $( \sqrt { 17 } , 0 ) , ( - \sqrt { 17 } , 0 )$; asymptotes $y = 4 x , y = - 4 x$
A) $y ^ { 2 } - \frac { x ^ { 2 } } { 16 } = 1$
B) $\frac { y ^ { 2 } } { 16 } - x ^ { 2 } = 1$
C) $\frac { x ^ { 2 } } { 16 } - y ^ { 2 } = 1$
D) $x ^ { 2 } - \frac { y ^ { 2 } } { 16 } = 1$

Question

Quizplus · Accepted Answer

A hyperbola centered at the origin with foci at  $( \sqrt { 17 } , 0 )$ and $( - \sqrt { 17 } , 0 )$ must have a horizontal transverse axis. This means that the standard form equation of the hyperbola is of the form

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

where $a$ is the distance from the center to each vertex and $b$ is the distance from the center to each co-vertex.

The distance between the foci is $2\sqrt{17}$. Since the graph has a horizontal transverse axis, the distance between the vertices is $2a$. Therefore,

$$2a = 2\sqrt{17} \quad \Rightarrow \quad a = \sqrt{17}$$

The slope of each asymptote is $\pm b/a = \pm 4$. Therefore, the asymptotes have slope $4$ and $-4$, and pass through the center of the hyperbola. Thus, the equations of the asymptotes are of the form

$$y = \pm 4x$$

Plugging in $y = 4x$ and $y = -4x$ into the equation of the hyperbola, we get

$$\frac{x^2}{17} - \frac{(4x)^2}{b^2} = 1 \quad \text{and} \quad \frac{x^2}{17} - \frac{(-4x)^2}{b^2} = 1$$

Simplifying each equation, we get

$$\frac{x^2}{17} - \frac{16x^2}{b^2} = 1 \quad \text{and} \quad \frac{x^2}{17} - \frac{16x^2}{b^2} = 1$$

Combining like terms and solving for $b$, we get

$$b^2 = 16\cdot 17 = 272 \quad \Rightarrow \quad b = 4\sqrt{17}$$

Thus, the equation of the hyperbola is

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

which is equivalent to choice D.

Find the Standard-Form Equation of the Hyperbola Centered at the Origin