Chapter 10, Problem 2RE

In Problems 1-10, identify each equation. If it is a parabola, give its vertex, focus, and directrix; if it is an ellipse, give its center, vertices, and foci: if it is a hyperbola, give its center, vertices, foci, and asymptotes.

$\frac{x^2}{25}-y^2\text{ = }1$

To determine

Each equation. If it is a parabola, give its vertex, focus, and directrix; if it is an ellipse, give its center, vertices, and foci; if it is a hyperbola, give its center, vertices, foci, and asymptotes.

Answer to Problem 2RE

Hyperbola

Center: $\left(0,0\right)$

Vertices: $\left(\pm 5,0\right)$

Foci: $\left(\pm \sqrt{26,0}\right)$

Asymptotes: $y=\pm \frac{1}{5}x$.

Explanation of Solution

Given:

$\frac{x^2}{25}-y²=1$

Formula used:

Equation	Center	Transverse axis	Foci	Vertices	Asymptotes
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ $b²=c²-a²$	$\left(0,0\right)$	$x\text{-axis}$	$\left(c,0\right)$	$\left(a,0\right)$	$y=\pm \frac{b}{a}x$
	$\left(0,0\right)$	$x\text{-axis}$	$\left(-c,0\right)$	$\left(-a,0\right)$

Calculation:

The equation $\frac{x^2}{25}-y^2=1$ is of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

It represents a hyperbola.

$a=5,b=1$

Center at origin $\left(0,0\right)$.

Vertices: $\left(\pm a,0\right)=\left(\pm 5,0\right)$.

$b²=c²-a²$

$c=\sqrt{25+1}=\surd 26$

Foci: $\left(\pm c,0\right)=\left(\pm \surd 26,0\right)$.

$y=\pm \frac{1}{5}x$