Find the standard-form equation of the hyperbola centered at the origin which satisfies the given conditions.
-Vertices at (0, 6) and (0, -6); foci at (0, 9) and (0, -9)
A) $\frac { y ^ { 2 } } { 45 } - \frac { x ^ { 2 } } { 36 } = 1$
B) $\frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { 81 } = 1$
C) $\frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { 45 } = 1$
D) $\frac { y ^ { 2 } } { 81 } - \frac { x ^ { 2 } } { 36 } = 1$

Question

Find the standard-form equation of the hyperbola centered at the origin which satisfies the given conditions.
-Vertices at (0, 6) and (0, -6); foci at (0, 9) and (0, -9) 
A) $\frac { y ^ { 2 } } { 45 } - \frac { x ^ { 2 } } { 36 } = 1$
B) $\frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { 81 } = 1$
C) $\frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { 45 } = 1$
D) $\frac { y ^ { 2 } } { 81 } - \frac { x ^ { 2 } } { 36 } = 1$

Quizplus · Accepted Answer

Since the center of the hyperbola is at the origin, the standard form equation of the hyperbola is either of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$ or $\frac{y^2}{a^2}-\frac{x^2}{b^2} = 1$.

We know that the distance between the center and vertices is 6, so $a=6$.

We also know that the distance between the center and foci is 9, so $c=9$.

Finally, we can use the relationship $a^2+b^2 = c^2$ for hyperbolas to find $b$. Substituting our known values, we get $6^2+b^2=9^2$. Solving for $b$, we get $b=3\sqrt{5}$.

Thus, the standard form equation of the hyperbola is $\frac{y^2}{36}-\frac{x^2}{45}=1$, which is equivalent to choice C.

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