Chapter 11, Problem 31RE

To determine

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

Answer to Problem 31RE

The given curve is a parabola with

The foci: (0,-2)

Vertices= (0,1)

Explanation of Solution

Given information:

The equation is $\frac{x^2}{12}+y=1$

Calculation:

For an quadratic equation ${\text{ax}}^{\text{2}}{\text{+bxy+cy}}^{\text{2}}\text{+dx+ey+f=0}$

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

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

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

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

Given equation is $\begin{array}{l}\frac{x^2}{12}+y=1\\ x^2+12y-12=0\end{array}$

Here;

A=1

B=0

C=0

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

So, the given equation is a parabola

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

General parabola equation is: $\begin{array}{l}{(}^x=4a(y-k)\\ \end{array}$

Here, h=0: k=I

And 4a=-12

a=-3

1. Vertices=(h,k)=(0,1)
2. Focus=(h,k+a)=(0,1-3)
=(0,-2)

  

Interpretation :

The given curve is a parabola with

The foci: (0,-2)

Vertices= (0,1)