Chapter 11, Problem 24RE

To determine

To find: The center, vertices, foci and asymptotes of the hyperbola & sketch of the hyperbola.

Answer to Problem 24RE

By converting the equation of hyperbola in general form it is calculated

Center of hyperbola: $\left(0,3\right)$

Vertices of hyperbola: $\left(0,6\right)\&\left(0,0\right)$

Foci of hyperbola: $\left(0,3\sqrt{2}+3\right)\&\left(0,-3\sqrt{2}+3\right)$

Asymptotes of hyperbola: $y=x\&y=-x$

Explanation of Solution

Given information:

Given hyperbola: $y^2=x^2+6y$

Calculation:

  $\frac{{\left(y-k\right)}^2}{b^2}-\frac{{\left(x-h\right)}^2}{a^2}=1$

  $\to$ Center $=\left(h,k\right)$

  $\to$ Vertices $=\left(h,\pm b+k\right)$

  $\to$ Foci $=\left(h,\pm be+k\right)$

  $\to$ Asymptotes $\Rightarrow y=\pm \frac{b}{a}x$

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

Or

  $\frac{{\left(y-k\right)}^2}{b^2}-\frac{{\left(x-h\right)}^2}{a^2}=1$

So, first convert given equation into general form

  $\begin{array}{l}\Rightarrow y^2=x^2+6y\\ \Rightarrow y^2-6y-x^2=0\\ \Rightarrow \left(y^2-6y+9\right)-9-x^2=0\left\{Adding\&subtracting\text{'}9\text{'}\right\}\\ \Rightarrow {\left(y-3\right)}^2-x^2=9\\ \Rightarrow \frac{{\left(y-3\right)}^2}{9}-\frac{{\left(x-0\right)}^2}{9}=1\end{array}$

Now compare it with general equation of hyperbola

Here, $\begin{array}{l}{a}^2=9\Rightarrow a=3\\ b^2=9\Rightarrow b=3\\ h=0\\ k=3\end{array}$

Now, center $=\left(h,k\right)=\left(0,3\right)$

Vertices $=\left(h,\pm b+k\right)=\left(0,\pm 3+3\right)$

(Adding & subtracting separately)

Vertices of hyperbola $=\left(0,6\right)\&\left(0,0\right)$

Foci $=\left(h,\pm be+k\right)$

  $=\left(h,\pm c+k\right)\left[\text{where; }c=be\right]$

Now, $c^2=a^2+b^2$

  $\begin{array}{l}{c}^2=9+9\\ c^2=18\\ c=\sqrt{18}\\ c=3\sqrt{2}\end{array}$

  $\Rightarrow$ Foci $\left(0,\pm 3\sqrt{2}+3\right)$

Foci $=\left(0,3\sqrt{2}+3\right)\&\left(0,-3\sqrt{2}+3\right)$

4. Asymptotes

  $\begin{array}{l}y=\pm \frac{b}{a}x\\ \Rightarrow y=\pm \frac{3}{3}x\\ y=\pm x\end{array}$

Asymptotes: $y=x\&y=-x$

  

Interpretation :

By converting the equation of hyperbola in general form it is calculated

Center of hyperbola: $\left(0,3\right)$

Vertices of hyperbola: $\left(0,6\right)\&\left(0,0\right)$

Foci of hyperbola: $\left(0,3\sqrt{2}+3\right)\&\left(0,-3\sqrt{2}+3\right)$

Asymptotes of hyperbola: $y=x\&y=-x$