Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin.
Vertices: (±3,0); passes through the point $( 5 , \sqrt { 3 } )$
A) $\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 27 / 16 } = - 1$
B) $\frac { x ^ { 2 } } { 27 / 16 } + \frac { y ^ { 2 } } { 9 } = 1$
C) $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 27 / 16 } = 1$
D) $\frac { x ^ { 2 } } { 27 / 16 } + \frac { y ^ { 2 } } { 9 } = - 1$
E) $\frac { y ^ { 2 } } { 9 } + \frac { x ^ { 2 } } { 27 / 16 } = 1$

Question

Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. 
Vertices: (±3,0); passes through the point  $( 5 , \sqrt { 3 } )$ 
A) $\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 27 / 16 } = - 1$
B) $\frac { x ^ { 2 } } { 27 / 16 } + \frac { y ^ { 2 } } { 9 } = 1$
C) $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 27 / 16 } = 1$
D) $\frac { x ^ { 2 } } { 27 / 16 } + \frac { y ^ { 2 } } { 9 } = - 1$
E) $\frac { y ^ { 2 } } { 9 } + \frac { x ^ { 2 } } { 27 / 16 } = 1$

Quizplus · Accepted Answer

The vertices at (±3,0) indicate that the major axis is along the x-axis, so the form of the equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $a = 3$. To find $b^2$, substitute the point $(5, \sqrt{3})$ into the equation, leading to $b^2 = \frac{27}{16}$.

Find the Standard Form of the Equation of the Hyperbola (5,3)( 5 , \sqrt { 3 } )(5,3​)

Find the Standard Form of the Equation of the Hyperbola $( 5 , \sqrt { 3 } )$