Chapter 10.4, Problem 35AYU

To determine

a.

To find: Center, transverse axis, vertices, foci and asymptotes.

Answer to Problem 35AYU

Solution:

Center: $(0,0)$.

The transverse axis is $y\text{-axis}$.

Vertex: $\left(0,5\right),\left(0,-5\right)$.

Foci: $\left(0,5\sqrt{2}\right)\text{and} \left(0,-5\surd 2\right)$.

Asymptotes: $y=x,y=-x$.

Explanation of Solution

Given:

$y^2-x^2=25$

Formula used:

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

Calculation:

Consider the given equation $y^2-x^2=25$.

Rewriting the equation $\frac{y^2}{25}-\frac{x^2}{25}=1$.

The center of the hyperbola is $(0,0)$.

From the given data we see that, transverse axis is $y\text{-axis}$.

$a=5,b=5,c=\sqrt{b^2+a^2},c=\sqrt{25+25}=5\surd 2$

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

Foci: $\left(0,\pm c\right)=\left(0,\pm 5\surd 2\right)$.

Asymptotes: $y=\pm \frac{a}{b}x=\pm x$.

To determine

b.

To graph: The equation $\frac{y^2}{25}-\frac{x^2}{25}=1$.

Answer to Problem 35AYU

Solution:

Explanation of Solution

Given:

$y^2-x^2=25$

Graph:

The graph of the equation of $\frac{y^2}{25}-\frac{x^2}{25}=1$ is plotted.

Using the given equation $y^2-x^2=25$. We find $a=5$, $b=5$. We have found the center: $(0,0)$.

The transverse axis is $y\text{-axis}$.

Vertex: $\left(0,\pm 5\right)$.

Foci: $\left(0,\pm 5\surd 2\right)$.

Asymptotes: $y=\pm x$.