Example: Graphing a Parabola with Vertex (0, 0) and the x-axis as the Axis of Symmetry

Answer: The standard form that applies to the given equation is [latex]{y}^{2}=4px[/latex]. Thus, the axis of symmetry is the x-axis. It follows that:

• [latex]24=4p[/latex], so [latex]p=6[/latex]. Since [latex]p>0[/latex], the parabola opens right the coordinates of the focus are [latex]\left(p,0\right)=\left(6,0\right)[/latex]
• the equation of the directrix is [latex]x=-p=-6[/latex]
• the endpoints of the focal diameter have the same x-coordinate at the focus. To find the endpoints, substitute [latex]x=6[/latex] into the original equation: [latex]\left(6,\pm 12\right)[/latex]

Example: Graphing a Parabola with Vertex (0, 0) and the y-axis as the Axis of Symmetry

Answer: The standard form that applies to the given equation is [latex]{x}^{2}=4py[/latex]. Thus, the axis of symmetry is the y-axis. It follows that:

• [latex]-6=4p[/latex], so [latex]p=-\frac{3}{2}[/latex]. Since [latex]p<0[/latex], the parabola opens down.
• the coordinates of the focus are [latex]\left(0,p\right)=\left(0,-\frac{3}{2}\right)[/latex]
• the equation of the directrix is [latex]y=-p=\frac{3}{2}[/latex]
• the endpoints of the focal diameter can be found by substituting [latex]\text{ }y=\frac{3}{2}\text{ }[/latex] into the original equation, [latex]\left(\pm 3,-\frac{3}{2}\right)[/latex]

Example: Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix

Answer: The focus has the form [latex]\left(p,0\right)[/latex], so the equation will have the form [latex]{y}^{2}=4px[/latex]. Multiplying [latex]4p[/latex], we have [latex]4p=4\left(-\frac{1}{2}\right)=-2[/latex]. Substituting for [latex]4p[/latex], we have [latex]{y}^{2}=4px=-2x[/latex]. Therefore, the equation for the parabola is [latex]{y}^{2}=-2x[/latex].

Example: Graphing a Parabola with Vertex (h, k) and Axis of Symmetry Parallel to the x-axis

Answer: 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]

Example: Graphing a Parabola from an Equation Given in General Form

Answer: 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]

• 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]

Example: Solving Applied Problems Involving Parabolas

1. Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane.
2. Use the equation found in part (a) to find the depth of the fire starter.

Cross-section of a travel-sized solar fire starter

Answer:

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-display]\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-display] The dish is about 0.74 inches deep.