Chapter A5.3, Problem 40E

(a)

To determine

To find: the counter examples to the given statements.

Answer to Problem 40E

Here, the curve is the case for a parabola and not an ellipse.

Explanation of Solution

Given: $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

  $AC>0$ , the graph is ellipse.

Calculation:

The given statement is false. The following example shows this:

Let $A=C=1,B=2$

  $AC>0$

Now,

  $\begin{array}{l}D=B^2-4AC\\ \text{ =}{\left(2\right)}^2-4\left(1\right)\left(1\right)\\ \text{ =0}\end{array}$

This is the case for a parabola, and not an ellipse.

Conclusion:

Hence, the curve is the case for a parabola and not an ellipse.

(b)

To determine

To find: the counter examples to the given statements.

Answer to Problem 40E

Here, the curve is the case for a parabola and not an hyperbola.

Explanation of Solution

Given:

  $AC>0$ , the graph is hyperbola.

Calculation:

Let $A=C=1,B=2$

  $AC>0$

Now,

  $\begin{array}{l}D=B^2-4AC\\ \text{ =}{\left(2\right)}^2-4\left(1\right)\left(1\right)\\ \text{ =0}\end{array}$

This is the case for a parabola, and not a hyperbola.

Conclusion:

Hence, the curve is the case for a parabola and not an hyperbola.

(c)

To determine

To find: the counter examples to the given statements.

Answer to Problem 40E

Here, the curve is definitely a hyperbola.

Explanation of Solution

Given:

  $AC<0$ , the graph is a hyperbola.

Calculation:

The given statement is true.

Given that $AC<0$

Now,

  $\begin{array}{l}AC<0\\ \Rightarrow 4AC<0\\ \Rightarrow -4AC>0\\ \Rightarrow B^2+\left(-4AC\right)>0\\ \Rightarrow B^2-4AC>0\\ \Rightarrow D>0\end{array}$

Since $D>0$ , the curve is definitely a hyperbola.

Conclusion:

Hence, the curve is definitely a hyperbola.