Find an equation for the hyperbola that has vertices $( \pm \sqrt { 7 } , 0 )$ and passes through the point $( - 4,6 )$
A) $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 7 } = 1$
B) $\frac { x ^ { 2 } } { 7 } - \frac { y ^ { 2 } } { 4 } = 1$
C) $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 28 } = 1$
D) $\frac { x ^ { 2 } } { 7 } - \frac { y ^ { 2 } } { 28 } = 1$
E) none of these

Question

Quizplus · Accepted Answer

The standard form of the equation of a hyperbola centered at the origin and with vertices on the x-axis is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, where $a$ is the distance from the center to a vertex along the x-axis and $b$ is the distance from the center to the positive y-intercept along the y-axis. In this case, the distance between the vertices is $2\sqrt{7}$, so $2a=2\sqrt{7}$ and $a=\sqrt{7}$. To find $b$, we note that the point $(0,b)$ is also on the hyperbola, so we can use the distance formula: $\sqrt{(0-\sqrt{7})^2+(b-0)^2}= \sqrt{(0+\sqrt{7})^2+(b-0)^2}$, which simplifies to $b^2=28$, so $b=2\sqrt{7}$.

Thus, the equation of the hyperbola is $\frac{x^2}{7}-\frac{y^2}{28}=1$. Plugging in the point $(-4,6)$, we find that $\frac{16}{7}-\frac{36}{28}=1$, so the point is on the hyperbola. Therefore, the answer is choice $	extbf{(D)}$.

Find an Equation for the Hyperbola That Has Vertices (±7,0)( \pm \sqrt { 7 } , 0 )(±7​,0)

Find an Equation for the Hyperbola That Has Vertices $( \pm \sqrt { 7 } , 0 )$