Write an equation for the hyperbola.
-$x$-intercepts $( \pm 5,0 )$; center at origin; passing through $( 5 \sqrt { 5 } , 6 )$
A) $\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 25 } = 1$

B) $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 9 } = 1$

C) $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 36 } = 1$

D) $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 25 } = 1$

Question

Write an equation for the hyperbola.
-$x$-intercepts $( \pm 5,0 )$; center at origin; passing through $( 5 \sqrt { 5 } , 6 )$
A) $\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 25 } = 1$

B) $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 9 } = 1$

C) $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 36 } = 1$

D) $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 25 } = 1$

Quizplus · Accepted Answer

The $x$-intercepts $(\pm 5,0)$ indicate that the vertices are on the $x$-axis, so the hyperbola opens left and right, and $a^2 = 25$. The point $(5\sqrt{5}, 6)$ lies on the hyperbola, so substituting $x = 5\sqrt{5}$ and $y = 6$ into the standard form of the hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and solving for $b^2$ gives $b^2 = 9$. Therefore, the correct equation is $\frac{x^2}{25} - \frac{y^2}{9} = 1$.

Write an Equation for the Hyperbola xxx -Intercepts (±5,0)( \pm 5,0 )(±5,0)

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