Chapter 11, Problem 32RE

To determine

To find: The type of curve; and its foci and vertices and its graph.

Answer to Problem 32RE

The given equation is an ellipse and the given curve is

  $\frac{x^2}{12}+\frac{y^2}{144}=\frac{y}{12}$

  $12x^2+y^2-12y=0$

  $\text{The foci:}$

  $(0,6+\sqrt{33})\&(0,6-\sqrt{33})$

  $\text{Vertices:}$

  $(\sqrt{3},6),(-\sqrt{3,6}),(0,12)\&(0,0)$

Explanation of Solution

Given information: Given equation; $12x^2+y^2-12y=0$

Calculation:

For an quadratic equation $a{x}^2+bxy+c{y}^2+dx+ey+f=0$

If $b^2-4ac=0$ - it is parabola

If $b^2-4ac>0$ - it is hyperbola

If $b^2-4ac<0$ and b=0 and a=c - it is circle

If $b^2-4ac<0$ and b $\ne$ 0 and a $\ne$ c - it is ellipse

Given equation is $12x^2+y^2-12y=0$

Here;

A=12

B=0

C=1

Now; $\begin{array}{l}{b}^2-4ac={(}^0-4(12)(1)\\ =-48\end{array}$

So, the given equation is an ellipse

  $\begin{array}{l}{x}^2+12y-12=0\\ x^2=-12(y-1)\end{array}$

Now, Focus:

For ellipse:

Focus: ( $h,k\pm c$ )

Centre: (h,k)

Vertices: ( $h\pm b,k$ ) and ( $h,k\pm a$ )

The general equation of ellipse: $\frac{{(x-h)}^2}{b^2}+\frac{{(y-k)}^2}{a^2}=1$

Given equation is:

  $\begin{array}{l}=12x^2+y^2-12y=0\\ =12x^2+[y^2-2(y)(6)+36-36]=0\\ =12x^2+{(}^y=26=36\\ =\frac{x^2}{3}+{\frac{(y-6)}{36}}^2=1\end{array}$

Here,

a=6

  $b=$

  $\sqrt{3}$

h=0

k=6

  $\begin{array}{l}{c}^2=a^2-b^2\\ c^2={(}^6+{(}^{\sqrt{3}}\\ c^2=33\\ c=\sqrt{33}\end{array}$

Vertices;

  $\begin{array}{l}=(h\pm b,k)\&(h,k\pm a)\\ =(0\pm \sqrt{3},6)\&(0,6\pm 6)\\ =(\sqrt{3},6),(-\sqrt{3},6)\&(0,12),(0,0)\end{array}$

Focus:

  $\begin{array}{l}=(h,k\pm c)\\ =(0,6\pm \sqrt{33})\\ =(0,6+\sqrt{33})\&(0,6-\sqrt{33})\end{array}$

  

Interpretation :

The given equation is an ellipse and the given curve is

  $\frac{x^2}{12}+\frac{y^2}{144}=\frac{y}{12}$

  $12x^2+y^2-12y=0$

  $\text{The foci:}$

  $(0,6+\sqrt{33})\&(0,6-\sqrt{33})$

  $\text{Vertices:}$

  $(\sqrt{3},6),(-\sqrt{3,6}),(0,12)\&(0,0)$