Chapter 13.5, Problem 51STP

To determine

To find: Whether the graph of each equation is parabola, circle, ellipse, or hyperbola.

Answer to Problem 51STP

Graph of given equation is Hyperbola

Explanation of Solution

Given information:

  $3x^2-2y^2+32y-134=0$

Concept:

Equation in the form

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Then we need to calculate discriminant

  $\Delta =B^2-4AC$

If $\Delta =0$ and $A=0$ or $C=0$

Then the figure is Parabola

If

  $\Delta >0$

Then the figure is Hyperbola

If $\Delta <0$ and $A=C$

Then the figure is Circle.

If $\Delta <0$ and $A\ne C$

Then the figure is Ellipse.

Given

  $3x^2-2y^2+32y-134=0$

On comparing with standard equation

We get

  $\begin{array}{l}A=3\\ B=0\\ C=-2\\ D=0\\ E=32\\ F=-134\end{array}$

Substituting values in discriminant formula

We get

  $\Rightarrow \Delta =0^2-4\left(3\right)\left(-2\right)=24>0$

Here

  $\Delta >0$

So, given equation is Hyperbola.