Chapter 11.5, Problem 31E

a.

To determine

To state: that whether the graph of the equation is a parabola, an ellipse or a hyperbola.

Answer to Problem 31E

The graph of the equation is hyperbola.

Explanation of Solution

Given information:

The equation is $6x^2+10xy+3y^2-6y=36$

Calculation: to find the discriminant,

  $\begin{array}{l}6x^2+10xy+3y^2-6y=36\\ A=6,B=10,C=3\end{array}$

Now, the discriminant is,

  $\begin{array}{l}{B}^2-4AC={\left(10\right)}^2-4\left(6\right)\left(3\right)\\ B^2-4AC=100-72\\ B^2-4AC>0\end{array}$

Thus, the graph of the equation is hyperbola.

b.

To determine

To graph: the equation by graphing utlility.

Explanation of Solution

Given information:

The equation is $6x^2+10xy+3y^2-6y=36$

Graph: Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $6x^2+10xy+3y^2-6y-36$ which is required to graph.

Step 4: Press GRAPH button to graph the function.

The graph is obtained as:

  

Interpretation: form the above graph it can be observed that the graph of equation is hyperbola.