Chapter 8, Problem 7RE

Solution

To find: The center, vertices and foci of the identified conic $\frac{x^2}{25}-\frac{y^2}{36}=1$ and sketch its graph.

Hyperbola, center is $\left(0,0\right)$ ,vertices is $\left(\pm 5,0\right)$ and foci is $\left(\pm \sqrt{61},0\right)$ .

Given information:

Equation of conic: $\frac{x^2}{25}-\frac{y^2}{36}=1$

Formula used:

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

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

Foci is $\left(\pm c,0\right)$

Vertices is $\left(\pm a,0\right)$

Pythagorean relation: $c^2=a^2+b^2$

Calculation:

The equation $\frac{x^2}{25}-\frac{y^2}{36}=1$ is of the form of standard equation of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Compare the equation with the standard equation of the hyperbola.

  $\begin{array}{c}{a}^2=25\\ a=5\\ b^2=36\\ b=6\end{array}$

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

Substitute $25$ for $a^2$ and $36$ for $b^2$ in the Pythagorean relation.

  $\begin{array}{c}{c}^2=25+36\\ =61\end{array}$

Take square root on both sides of the equation.

  $c=\sqrt{61}$

Substitute $\sqrt{61}$ for $c$ in the foci formula.

Therefore, foci is $\left(\pm \sqrt{61,0}\right)$ .

Substitute $4$ for $a$ in the vertices formula.

Therefore, vertices is $\left(\pm 5,0\right)$ .

Draw the graph as follows:

  

Hence, the given equation of conic is a hyperbola, center is $\left(0,0\right)$ ,vertices is $\left(\pm 5,0\right)$ and foci is $\left(\pm \sqrt{61},0\right)$ .