Chapter A5.3, Problem 43E

(a)

To determine

To describe: the representation of the equation.

Answer to Problem 43E

Here, the given curve represents a parabola.

Explanation of Solution

Given: $x^2+4xy+4y^2+6x+12y+9=0$ and $2y=-x-3$ .

Calculation:

Given the following equation:

  $x^2+4xy+4y^2+6x+12y+9=0$

Comparing with the standard equation for a Quadratic Curve,

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Thus,

  $A=1,\text{ B=4, C=4}$

Now, discriminate $\text{D}$

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

Since $\text{D=0}$ , the given curve represents a parabola.

Conclusion:

Hence, the given curve represents a parabola.

(b)

To determine

To show: The graph of the equation is a given line.

Answer to Problem 43E

Here, the curve is a straight line.

Explanation of Solution

Given:

Line $2y=-x-3$ .

Calculation:

The left hand side of the given equation factors as a perfect square.

  $x^2+4xy+4y^2+6x+12y+9={\left(x+2y+3\right)}^2$

Therefore, the equation becomes

  $\begin{array}{l}{\left(x+2y+3\right)}^2=0\\ \Rightarrow x+2y+3=0\\ \Rightarrow 2y=-x-3\end{array}$

So, the curve is a degenerate parabola (i.e. a straight line).

Conclusion:

Hence, the curve is a degenerate parabola which is a straight line.