Chapter 9, Problem 2CT

To determine

To find: Vertices and Foci of given conic.

Answer to Problem 2CT

Conic is Ellipse.

Foci of ellipse are $\left(12\pm 2\sqrt{21},0\right)$

Vertices of Ellipse are $\left(12\pm 4\sqrt{11},0\right)$ .

Explanation of Solution

Given information:

  $x^2+4y^2-24x+32=0$

  $x^2+4y^2-24x+32=0$

Graph of given Conic is

  

Here red line represents the given conic

So, from the graph we can say that given conic is Ellipse.

  $\begin{array}{l}{x}^2+4y^2-24x+32=0\\ \Rightarrow x^2-2\left(12\right)\left(x\right)+{\left(12\right)}^2+4y^2+32={\left(12\right)}^2\\ \Rightarrow {\left(x-12\right)}^2+4y^2=144-32\\ \Rightarrow {\left(x-12\right)}^2+4y^2=112\\ \Rightarrow \frac{{\left(x-12\right)}^2}{112}+\frac{4y^2}{112}=1\\ \Rightarrow \frac{{\left(x-12\right)}^2}{112}+\frac{y^2}{28}=1\end{array}$

If Ellipse in the form

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

Then

Centre of Ellipse is $\left(h,k\right)$

Coordinates of Vertices of Ellipse are $\left(h\pm a,k\right)$

Coordinates of Foci of Ellipse are $\left(h\pm c,k\right)$

Where $c^2=a^2-b^2$

  $\frac{{\left(x-12\right)}^2}{112}+\frac{y^2}{28}=1$

Comparing with general equation

We get

  $\begin{array}{l}h=12\\ k=0\\ a^2=112\\ b^2=28\end{array}$

Therefore

  $\begin{array}{l}{c}^2=a^2-b^2\\ \Rightarrow c^2=112-28\\ \Rightarrow c^2=84\\ \Rightarrow c=\sqrt{84}\\ \Rightarrow c=2\sqrt{21}\end{array}$

Therefore

Foci of Ellipse = $\left(h\pm c,k\right)$

Foci of ellipse are $\left(12\pm 2\sqrt{21},0\right)$

  $\begin{array}{l}{a}^2=176\\ \Rightarrow a=\sqrt{176}\\ \Rightarrow a=4\sqrt{11}\end{array}$

Vertices of Ellipse = $\left(h\pm a,k\right)$

Vertices of Ellipse are $\left(12\pm 4\sqrt{11},0\right)$ .