Chapter 8.6, Problem 42E

Solution

To find: The type of conic section, write equation in standard form and graph of the equation.

equation $y^2+14y+16x+33=0$ is a parabola and the standard form is ${\left(y+7\right)}^2=-16\left(x-1\right)$ . The graph is shown in figure (1).

Given information: The given equation is $y^2+14y+16x+33=0$.

Formula used: The general second-degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ describe any conic.

The type of conic is a circle if $B^2-4AC<0,B=0,A=C$.

The type of conic is an ellipse if $B^2-4AC<0,B\ne 0,A\ne C$.

The type of conic is a parabola if $B^2-4AC=0$.

The type of conic is a hyperbola if $B^2-4AC>0$.

Calculation:

Consider the given equation.

  $y^2+14y+16x+33=0$

Compare the above equation with the general equation of the conic.

  $A=0,B=0,C=1$

Substitute $0$ for $A$ , $0$ for $B$ and $1$ for $C$ in $B^2-4AC$ to find the discriminant.

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

Since $B^2-4AC=0$ . So, the equation satisfies all the conditions of a parabola.

Rewrite the given equation in the standard form.

  $\begin{array}{c}{y}^2+14y=16x-33\\ \left(y^2+14y+49\right)-49=-16x-33\\ {\left(y+7\right)}^2=-16x+16\\ {\left(y+7\right)}^2=-16\left(x-1\right)\end{array}$

The vertex of the parabola is at $\left(1,-7\right)$.

Drawn the graph of the parabola using the above conditions.

  

Figure (1)

Therefore, the equation $y^2+14y+16x+33=0$ is a parabola and the standard form is ${\left(y+7\right)}^2=-16\left(x-1\right)$.