Brandyn is graphing parabolas as a part of his homework assignment. He is familiar with the standard form of a parabola: \begin{align*}y = ax^2\end{align*}

On his third problem, he runs into a snag. He has simplified the equation significantly, and has been trying to get it to fit the standard form, but he keeps coming up with this: \begin{align*}x - 4 = 3(y - 3)^2\end{align*}*y* term is the squared one, instead of the *x* term.

What is going on here?

### Parabolas and Analytic Geometry

This is our second lesson on parabolas. In the initial lesson, we explored the parabola using the distance formula, and touched on the use of the focus and directrix. In this lesson, we first examine parabolas from the "analytic geometry" point of view, and then work a few examples with the focus and directrix of a parabola.

#### Finding the Equation of a Parabola Using Analytic Geometry

Consider a cone oriented in space as pictured below:

If the cone opens at an angle such that at any point its radius to height ratio is \begin{align*}a\end{align*}

\begin{align*}\sqrt{(x - 0)^2 + (y - 0)^2} = az\end{align*}

Or:

\begin{align*}x^2 + y^2 = a^2 z^2\end{align*}

This equation works for negative values of \begin{align*}x, y\end{align*}

To consider the intersection of this cone with a plane that is parallel to the line marked \begin{align*}l\end{align*}

In the above diagram, \begin{align*}P(x,z)\end{align*}

\begin{align*}\cos(90 - \alpha - \theta) = \sin(\alpha + \theta),\end{align*}

and substituting with the sine addition formula gives us:

\begin{align*}\frac{SQ'}{\sqrt{x^2 + z^2}} = \sin(\alpha) \cos(\theta) + \cos(\alpha) \sin(\theta),\end{align*}

which we can use our diagram to change to:

\begin{align*}\frac{SQ'}{\sqrt{x^2 + z^2}} = \frac{x}{\sqrt{x^2 + z^2}} \cos(\theta) + \frac{z}{\sqrt{x^2 + z^2}} \sin(\theta)\end{align*}

which simplifies to:

\begin{align*}SQ' = x \cos(\theta) + z \sin(\theta)\end{align*}

To find the \begin{align*}x-\end{align*}coordinates of our rotated point \begin{align*}P'\end{align*}, we can use the fact that \begin{align*}\sin(90 - \alpha - \theta) = \frac{P' Q'}{\sqrt{x^2 + z^2}}\end{align*}. But by properties of sine we have:

\begin{align*}\sin(90 - \alpha - \theta) = \cos(\alpha + \theta)\end{align*}

and substituting with the cosine addition formula gives us:

\begin{align*}\frac{P' Q'}{\sqrt{x^2 + z^2}} = \cos(\alpha) \cos(\theta) - \sin(\alpha)\sin(\theta),\end{align*}

which we can use our diagram to change to:

\begin{align*}\frac{P' Q'}{\sqrt{x^2 + z^2}} = \frac{z}{\sqrt{x^2 + z^2}} \cos(\theta) - \frac{x}{\sqrt{x^2 +z^2}} \sin(\theta)\end{align*}

which simplifies to:

\begin{align*}P'Q'=z \cos(\theta) - x \sin(\theta)\end{align*}

Looking back at the picture, this means that the coordinates of \begin{align*}P'\end{align*} are \begin{align*}(x \cos(\theta) + z \sin(\theta), \ z \cos(\theta) - x \sin(\theta))\end{align*}. In other words, in rotating from \begin{align*}P\end{align*} to \begin{align*}P'\end{align*}, the \begin{align*}x-\end{align*}coordinate changes to \begin{align*}x \cos(\theta) + z \sin(\theta)\end{align*} and the \begin{align*}z-\end{align*} coordinate changes to \begin{align*}z \cos(\theta) - x \sin(\theta)\end{align*}.

If this rotation happens to every point on the cone, we can substitute \begin{align*}x \cos(\theta) + z \sin(\theta)\end{align*} for \begin{align*}x\end{align*} and \begin{align*}z \cos(\theta) - x \sin(\theta)\end{align*} for \begin{align*}z\end{align*} into our equation of the cone, resulting in a new equation for the cone after rotating by \begin{align*}\theta\end{align*}.

\begin{align*}(x \cos(\theta) + z \sin(\theta))^2 + y^2 & = a^2(z \cos(\theta) - x \sin(\theta))^2 \\ x^2 \cos^2(\theta) + 2xz \cos(\theta) \sin(\theta) + z^2 \sin^2(\theta) + y^2 & = a^2(x^2 \sin^2(\theta) - 2xz \cos(\theta) \sin(\theta) + z^2 \cos^2(\theta)) \\ x^2 \cos^2(\theta) + 2xz \cos(\theta) \sin(\theta) + z^2 \sin^2(\theta) + y^2 & = a^2 x^2 \sin^2(\theta) - 2a^2 xz \cos(\theta) \sin(\theta) + a^2 z^2 \cos^2(\theta)) \\ x^2 \cos^2(\theta) + 2xz \cos(\theta) \sin(\theta) + z^2 \sin^2(\theta) + y^2 & = a^2 x^2 \sin^2(\theta) - 2a^2 xz \cos(\theta) \sin(\theta) + a^2 z^2 \cos^2(\theta))\end{align*}

Now in the case of the tilted cone, we want to tilt the cone such that the left side becomes vertical. Since the factor \begin{align*}a\end{align*} determines how tilted the cone is, we can see from the triangle below that \begin{align*}\sin(\theta) = \frac{a}{\sqrt{1 + a^2}}\end{align*} and \begin{align*}\cos(\theta) = \frac{1}{\sqrt{1 + a^2}}\end{align*}.

So the equation becomes:

\begin{align*}x^2 \frac{1}{1 + a^2} + 2xz \frac{a}{1 + a^2} + z^2 \frac{a^2}{1 + a^2} + y^2 & = a^2 x^2 \frac{a^2}{1 + a^2} - 2a^2 xz \frac{a}{1 + a^2} + a^2 z^2 \frac{1}{1 + a^2} \\ x^2 + 2xza + z^2 a^2 + y^2(1 + a^2) & = a^4 x^2 - 2a^3xz + a^2 z^2 \\ x^2 + 2xza + y^2(1 + a^2) & = a^4 x^2 - 2a^3 xz\end{align*}

Now that we have tilted our cone, to take a cross section that is parallel to the left side of the cone, we can simply cut it with a vertical plane. The equation of a vertical plane going through \begin{align*}(b,0,0)\end{align*} and perpendicular to the \begin{align*}x-\end{align*}axis is \begin{align*}x = b\end{align*}. Therefore, setting \begin{align*}x\end{align*} equal to the constant in the equation above will give us the intersection of the tilted cone and a plane parallel to one side of the cone. \begin{align*}b\end{align*}. Here is a picture of the rotation and the cross-section, which lies in an \begin{align*}xz-\end{align*}plane.

Setting \begin{align*}x\end{align*} equal to the constant \begin{align*}b\end{align*}, we have:

\begin{align*}b^2 + 2abz + y^2(1 + a^2) & = a^4 b^2 - 2a^3 bz \\ z(2ab + 2a^3 b) & = -y^2(1 + a^2) +a^4 b^2 - b^2 \\ z & =\left (\frac{-1 - a^2}{2ab + 2a^3 b}\right )y^2 + (a^4 b^2 - b^2)\end{align*}

Although this coefficient and constant term seem complicated, \begin{align*}a\end{align*} and \begin{align*}b\end{align*} can be chosen so that the coefficient of the \begin{align*}y^2\end{align*} term can be equal to any number (you will explore this fact in an exercise). The constant term can be ignored since any parabola can be shifted vertically by any amount.

So the general form of a parabola is:

\begin{align*}z = Ay^2\end{align*}

where \begin{align*}A\end{align*} is any constant.

Or, using the more standard \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates the form of a parabola is

\begin{align*}y = ax^2\end{align*}

As before, this equation can be adapted to produce the shifted and horizontally oriented forms.

### Examples

#### Example 1

Earlier, you were asked a question about Brandyn, who is unsure why he has a y^{2} term in his standard form equation instead of an x^{2} term.

Brandyn keeps coming up with a *y*^{2} term because this is a sideways parabola.

#### Example 2

In the proof above, we greatly simplified the formula near the end by substituting *A* for \begin{align*}\frac{-1 -a^2}{2ab + 2a^3 b}\end{align*}.

Explain why this was permissible by showing that for any \begin{align*}A\end{align*} there exist constants \begin{align*}a\end{align*} and \begin{align*}b\end{align*} such that \begin{align*}A = \frac{-1 -a^2}{2ab + 2a^3 b}\end{align*}.

Solving for \begin{align*}b\end{align*} in terms of \begin{align*}A\end{align*} and \begin{align*}a\end{align*}, we have: \begin{align*}A(2ab + 2a^3 b) & = -1 - a^2 \\ 2Aab(1 + a^2) & = -(1 +a^2) \\ 2 Aab & = -1 \\ 2Aab & = -1 \\ b & = -\frac{1}{2Aa}\end{align*} So we can set \begin{align*}b=2Aa\end{align*} and the relationship will hold.

#### Example 3

Draw a sketch of the following parabola. Also identify its directrix and focus. \begin{align*}3x^2 + 6x - y = 0\end{align*}

Factor and complete the square to get: \begin{align*}3(x + 1) = y + 3\end{align*}. The **vertex** is at: \begin{align*}(-1, -3)\end{align*}.

The focus is \begin{align*}(-1, -2 \frac{11}{12})\end{align*} The directrix is the line \begin{align*}y = -3 \frac{1}{12}\end{align*}

#### Example 4

Find the equation for a parabola with directrix \begin{align*}y=-2\end{align*} and focus (3,8).

The vertex is vertically midway between the focus and directrix: \begin{align*}\frac {-2 + 8}{2} = 3\end{align*}, the same horizontally as the focus: \begin{align*}x = 3\end{align*} and therefore at: \begin{align*}(3, 3)\end{align*}.

Substituting those values into the formula gives:

\begin{align*}y - 3 = \frac{1}{20}(x -3)^2\end{align*}

#### Example 5

Find the equation for a parabola with directrix \begin{align*}y = 3\end{align*} and focus (2, -1).

Using the vertex form of a parabola \begin{align*}(y - k)^2 = 4a(x - h)\end{align*}:

Recall that the vertex y-value \begin{align*}k\end{align*} is the midpoint of the directrix and the focus on the line perpendicular to the directrix and crossing the focus. Therefore the y-value of the vertex is \begin{align*}1\end{align*}

Recall that the vertex x-value \begin{align*}h\end{align*} is the same as the focus, therefore the vertex x-value is \begin{align*}2\end{align*}

Finally, recall that \begin{align*}a\end{align*} the distance from the vertex to the focus or from the vertex to the directrix (which are the same): \begin{align*}\therefore a = 2\end{align*}

Substituting gives:\begin{align*}(y - 1)^2 = 8(x - 2) \to y^2 +1 = 8x -16 \to y^2 = 8x -17\end{align*}

#### Example 6

Describe the shape of a parabola as it relates to a cone or double cone.

The shape of a parabola as it relates to a cone or double cone, is that a parabola represents the revealed shape when a hollow cone is sliced through at an angle equal to the side of the cone. Particularly clear with a double cone is the fact that slicing through at a steeper angle will result in two curves (a hyperbola) and a shallower angle will result in an ellipse.

#### Example 7

Sketch the following parabola and identify the directrix and focus: \begin{align*}4x^2 -3x +y = 7\end{align*}.

\begin{align*}y = ax^2 + bx + c\end{align*} ..... Recall the standard form of a parabola

\begin{align*}a = -3 | b = 6 | c = -2\end{align*} ..... Extract a, b, c

\begin{align*}x = \frac{-6}{2(-3)} \to x = 1\end{align*} ..... The x-coordinate of the vertex = \begin{align*}\frac{-b}{2a}\end{align*}

\begin{align*}y = (-3)(1) + 6(1) -2\end{align*} ..... Substitute the calculated x-value to solve for y

\begin{align*}y = 1\end{align*} ...... The vertex = \begin{align*}(1, 1)\end{align*}

\begin{align*}x = 1\pm\frac{1}{\sqrt{3}}\end{align*} ..... Identify the x-intercepts using the Quadratic Formula

### Review

Use the image to identify the parts of the parabola:

- The Focus
- The Vertex
- The Focal Radius
- The Directrix
- The Parabola

Use the image *and given equation* of the parabola to identify the following:

- The coordinates of the focus
- The equation of the directrix
- The length of the focal radius
- The equation of the axis of symmetry
- The coordinates of the vertex

- Find the equation for a parabola with directrix:\begin{align*}x = 2\end{align*} and focus: \begin{align*}(0, -2)\end{align*}
- Find the equation for a parabola with vertex:\begin{align*}(5, -2)\end{align*} and directrix:\begin{align*} y = -5\end{align*}
- Find the equation for a parabola with focus:\begin{align*}(3, 5)\end{align*} and vertex:\begin{align*}(3, 1)\end{align*}

Use the image to identify the vertex, axis of symmetry and equation of the parabola:

### Review (Answers)

To see the Review answers, open this PDF file and look for section 6.4.

### Resources