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, and the distance from the center to a vertex is 3, so $a^2 = 9$. Since the hyperbola passes through $(5, \sqrt{3})$, we can use this point to find $b^2$. Substituting $x = 5$ and $y = \sqrt{3}$ into the standard form of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we get $\frac{25}{9} - \frac{3}{b^2} = 1$. Solving for $b^2$ gives $b^2 = \frac{27}{16}$. Therefore, the correct equation is $\frac{x^2}{9} - \frac{y^2}{27/16} = 1$.

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 $( 5 , \sqrt { 3 } )$