Chapter 8, Problem 6RE

Solution

To find: The center, vertices and foci of the identified conic $\frac{y^2}{16}-\frac{x^2}{49}=1$ and sketch its graph.

Hyperbola; Center: $\left(0,0\right)$ ;vertices is $\left(0,\pm 4\right)$ ; foci: $\left(0,\pm \sqrt{65}\right)$ .

Given information:

Equation of conic: $\frac{y^2}{16}-\frac{x^2}{49}=1$

Formula used:

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

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

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

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

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

Calculation:

The equation $\frac{y^2}{16}-\frac{x^2}{49}=1$ is of the form of standard equation of the hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

Compare the equation with the standard equation of the hyperbola.

  $\begin{array}{c}{a}^2=16\\ a=4\\ b^2=49\\ b=7\end{array}$

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

Substitute $16$ for $a^2$ and $49$ for $b^2$ in the Pythagorean relation.

  $\begin{array}{c}{c}^2=16+49\\ =65\end{array}$

Take square root on both sides of the equation.

  $c=\sqrt{65}$

Substitute $\sqrt{65}$ for $c$ in the foci formula.

Therefore, foci is $\left(0,\pm \sqrt{65}\right)$ .

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

Therefore, vertices is $\left(0,\pm 4\right)$ .

Draw the graph as follows:

  

Hence, the given equation of conic is a hyperbola, center is $\left(0,0\right)$ ,vertices is $\left(0,\pm 4\right)$ and foci is $\left(0,\pm \sqrt{65}\right)$ .