Chapter A5.3, Problem 44E

(a)

To determine

To describe: the representation of the equation.

Answer to Problem 44E

Here, the given equation represents a parabola.

Explanation of Solution

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

Calculation:

Given Equation:

  $9x^2+6xy+y^2-12x-4y+4=0$

To decide whether the given equation represents a parabola, an ellipse, or a hyperbola.

Comparing with the standard equation for a Quadratic Curve,

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Thus,

  $A=9,\text{ B=6, C=1}$

Now, discriminate $\text{D}$

  $\begin{array}{l}{\text{=B}}^2-4AC\\ ={\left(6\right)}^2-4\left(9\right)\left(1\right)\\ =0\end{array}$

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

Conclusion:

Hence, the given equation represents a parabola.

(b)

To determine

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

Answer to Problem 44E

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

Explanation of Solution

Given:

line $y=-3x+2$ .

Calculation:

The left hand side of the given equation factors as a perfect square. $9x^2+6xy+y^2-12x-4y+4={\left(3x+y-2\right)}^2$

Therefore, the equation becomes

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

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

Conclusion:

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