Write Equations of Hyperbolas in Standard Form
-Foci: (-5, 0), (5, 0); vertices: (-2, 0), (2, 0)
A) $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 21 } = 1$
B) $\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 21 } = 1$
C) $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 25 } = 1$
D) $\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1$

Question

Write Equations of Hyperbolas in Standard Form
-Foci: (-5, 0), (5, 0); vertices: (-2, 0), (2, 0) 
A) $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 21 } = 1$
B) $\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 21 } = 1$
C) $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 25 } = 1$
D) $\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1$

Quizplus · Accepted Answer

The standard form of a hyperbola with a horizontal transverse axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where 2a is the distance between the vertices, and 2c is the distance between the foci. Here, a = 2 (since the vertices are at (-2, 0) and (2, 0)), so $a^2 = 4$. The distance between the foci is 10 (from (-5, 0) to (5, 0)), so $c = 5$ and $c^2 = 25$. Using $c^2 = a^2 + b^2$, we find $b^2 = c^2 - a^2 = 25 - 4 = 21$. Therefore, the equation is $\frac{x^2}{4} - \frac{y^2}{21} = 1$.

Write Equations of Hyperbolas in Standard Form -Foci: (-5, 0) x24−y221=1\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 21 } = 14x2​−21y2​=1

Write Equations of Hyperbolas in Standard Form
-Foci: (-5, 0) $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 21 } = 1$