The standard form that applies to the given equation is [latex]{\left(y-k\right)}^{2}=4p\left(x-h\right)[/latex]. Thus, the axis of symmetry is parallel to the x-axis. It follows that:

• the vertex is [latex]\left(h,k\right)=\left(-3,1\right)[/latex]
• the axis of symmetry is [latex]y=k=1[/latex]
• [latex]-16=4p[/latex], so [latex]p=-4[/latex]. Since [latex]p<0[/latex], the parabola opens left.
• the coordinates of the focus are [latex]\left(h+p,k\right)=\left(-3+\left(-4\right),1\right)=\left(-7,1\right)[/latex]
• the equation of the directrix is [latex]x=h-p=-3-\left(-4\right)=1[/latex]
• the endpoints of the focal diameter are [latex]\left(h+p,k\pm 2p\right)=\left(-3+\left(-4\right),1\pm 2\left(-4\right)\right)[/latex], or [latex]\left(-7,-7\right)[/latex] and [latex]\left(-7,9\right)[/latex]

Next we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.

Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is [latex]{\left(x-h\right)}^{2}=4p\left(y-k\right)[/latex]. Thus, the axis of symmetry is parallel to the y-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable [latex]x[/latex] in order to complete the square.

[latex]\begin{array}{l}{x}^{2}-8x - 28y - 208=0\hfill \\ \text{ }{x}^{2}-8x=28y+208\hfill \\ \text{ }{x}^{2}-8x+16=28y+208+16\hfill \\ \text{ }{\left(x - 4\right)}^{2}=28y+224\hfill \\ \text{ }{\left(x - 4\right)}^{2}=28\left(y+8\right)\hfill \\ \text{ }{\left(x - 4\right)}^{2}=4\cdot 7\cdot \left(y+8\right)\hfill \end{array}[/latex]

It follows that:

• the vertex is [latex]\left(h,k\right)=\left(4,-8\right)[/latex]
• the axis of symmetry is [latex]x=h=4[/latex]
• since [latex]p=7,p>0[/latex] and so the parabola opens up
• the coordinates of the focus are [latex]\left(h,k+p\right)=\left(4,-8+7\right)=\left(4,-1\right)[/latex]
• the equation of the directrix is [latex]y=k-p=-8 - 7=-15[/latex]
• the endpoints of the focal diameter are [latex]\left(h\pm 2p,k+p\right)=\left(4\pm 2\left(7\right),-8+7\right)[/latex], or [latex]\left(-10,-1\right)[/latex] and [latex]\left(18,-1\right)[/latex]

Next we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.

1. The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form [latex]{x}^{2}=4py[/latex], where [latex]p>0[/latex]. The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have [latex]p=1.7[/latex].
   [latex]\begin{array}{ll}{x}^{2}=4py\hfill & \begin{array}{cccc}& & & \end{array}\text{Standard form of upward-facing parabola with vertex (0,0)}\hfill \\ {x}^{2}=4\left(1.7\right)y\hfill & \begin{array}{cccc}& & & \end{array}\text{Substitute 1}\text{.7 for }p.\hfill \\ {x}^{2}=6.8y\hfill & \begin{array}{cccc}& & & \end{array}\text{Multiply}.\hfill \end{array}[/latex]
2. The dish extends [latex]\frac{4.5}{2}=2.25[/latex] inches on either side of the origin. We can substitute 2.25 for [latex]x[/latex] in the equation from part (a) to find the depth of the dish.
   [latex]\begin{array}{ll}\text{ }{x}^{2}=6.8y\hfill & \text{Equation found in part (a)}.\hfill \\ {\left(2.25\right)}^{2}=6.8y\hfill & \text{Substitute 2}\text{.25 for }x.\hfill \\ \text{ }y\approx 0.74 \hfill & \text{Solve for }y.\hfill \end{array}[/latex]
   The dish is about 0.74 inches deep.