Find the vertices of the hyperbola. $16 y ^ { 2 } - 4 x ^ { 2 } = 64$
A) $( 2,0 )$
B) $( 0 , \pm 1 )$
C) $( \pm 2,0 )$
D) $( 0 , \pm 2 )$
E) none of these

Question

Quizplus · Accepted Answer

The given equation is $16y^2 - 4x^2 = 64$, which can be rewritten as $\frac{y^2}{4} - \frac{x^2}{16} = 1$. This is the standard form of a hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, where $a^2 = 4$ and $b^2 = 16$. The vertices of a hyperbola in this form are at $(0, \pm a)$, so in this case, the vertices are at $(0, \pm 2)$.

Find the Vertices of the Hyperbola 16y2−4x2=6416 y ^ { 2 } - 4 x ^ { 2 } = 6416y2−4x2=64

Find the Vertices of the Hyperbola $16 y ^ { 2 } - 4 x ^ { 2 } = 64$