Chapter 8.5, Problem 10QR

Solution

To find: The focii and vertices of the conic.

The focii is $\left(0,\pm 2\sqrt{10}\right)$ and vertices of the conic is $\left(0,\pm 6\right)$ .

Given information:

The equation of conic is $\frac{y^2}{36}-\frac{x^2}{4}=1$ .

Formula used:

The general equation of conic is $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ .

The coordinates of the vertices are $\left(h,k\pm a\right)$ .

The formula to calculate $c$ is $c^2=a^2+b^2$ .

The location of focii are $\left(h,k\pm c\right)$ .

Calculation:

Compare the general equation of conic $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ with $\frac{y^2}{36}-\frac{x^2}{4}=1$ .

The value of $h=0$ , $k=0$ .

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

And $\begin{array}{c}{b}^2=4\\ b=2\end{array}$

Substitute $6$ for $a$ and $2$ for $b$ in $c^2=a^2+b^2$ .

  $\begin{array}{c}{c}^2=a^2+b^2\\ =6^2+2^2\\ c=2\sqrt{10}\end{array}$

Substitute $0$ for $h$ , $0$ for $k$ , and $2\sqrt{10}$ for $c$ in location of focii $\left(h,k\pm c\right)$ .

  $\begin{array}{l}\left(h,k+c\right)=\left(0,2\sqrt{10}\right)\\ \left(h,k-c\right)=\left(0,-2\sqrt{10}\right)\end{array}$

Substitute $0$ for $h$ , $0$ for $k$ , and $6$ for $a$ in location of vertices $\left(h,k\pm a\right)$ .

  $\begin{array}{l}\left(h,k+a\right)=\left(0,6\right)\\ \left(h,k-c\right)=\left(0,-6\right)\end{array}$

Therefore, focii is $\left(0,\pm 2\sqrt{10}\right)$ and vertices of the conic is $\left(0,\pm 6\right)$ .