How does the shape of a parabola change as its focus and directrix move further apart?

### Equations of Parabolas

Consider a point and a line.

Now, consider all the points that are the same distance away from the point as the line. Below are four such points:

Can you imagine what the set of ALL the points would look like? The set of all points is a parabola:

In the past you thought of a parabola in terms of its vertex and intercepts. Another way of defining a **parabola** is as *the set of points that are equidistant from a point* *called the* *focus**and a line* *called the* ** directrix**. This alternate way of thinking about a parabola leads to another general equation for parabolas:

- For parabolas opening up or down with a horizontal directrix: \begin{align*}(x-h)^2=4p(y-k) \end{align*}
- For parabolas opening left or right with a vertical directrix: \begin{align*}(y-k)^2=4p(x-h)\end{align*}

In each case, \begin{align*}(h, k) \end{align*} is the vertex of the parabola. \begin{align*}p\end{align*} is the distance from the vertex to the focus (and also the distance from the vertex to the directrix.

To derive the general equation for the parabola, pick a generic point on a parabola and find an equation that shows the relationship between the distance from that point to the focus and the distance from that point to the directrix.

#### Deriving the Equation of a Parabola

1. Consider the parabola below. The vertex is at \begin{align*}(h, k) \end{align*} and the distance from the focus to the vertex is length \begin{align*}p\end{align*}. What is the equation of the directrix in terms of \begin{align*}h\end{align*}, \begin{align*}k\end{align*}, and \begin{align*}p\end{align*}? At what point is the focus in terms of \begin{align*}h\end{align*}, \begin{align*}k\end{align*}, and \begin{align*}p\end{align*}?

Because the directrix is also a distance of \begin{align*}p\end{align*} from the vertex, it must be at height \begin{align*}k-p\end{align*}. The equation of the directrix is \begin{align*}y=k-p\end{align*}.

The focus has the same \begin{align*}x\end{align*}-coordinate as the vertex and has a height that is \begin{align*}p\end{align*} more than the height of the vertex. The focus is at \begin{align*}(h, k+p)\end{align*}.

2. Consider a random point \begin{align*}(x, y) \end{align*} on the parabola connected to the focus with a right triangle.

Find the distance from the point to the focus using the Pythagorean Theorem.

The two legs of the triangle are \begin{align*}x-h \end{align*} and \begin{align*}y-(k+p)\end{align*}. The distance from the focus to the point is the hypotenuse of the triangle, \begin{align*}c\end{align*}.

\begin{align*}& (x-h)^2+(y-(k+p))^2=c^2 \\ & c=\sqrt{(x-h)^2+(y-(k+p))^2} \end{align*}

Therefore, the distance from the point to the focus is \begin{align*}\sqrt{(x-h)^2+(y-(k+p))^2}\end{align*}.

3. Find the perpendicular distance from the point to the directrix.

The distance from the point to the directrix is \begin{align*}y-(k-p)\end{align*}.

Use the two distances from the previous problem to derive the equation \begin{align*}(x-h)^2=4p(y-k)\end{align*}. Why does this have to be the equation of the parabola?

You are assuming that the random point \begin{align*}(x, y)\end{align*} is on the parabola, and therefore it must be the same distance from the focus as from the directrix. This means that the two distances from the previous problem must be equivalent.

\begin{align*}\sqrt{(x-h)^2+(y-(k+p))^2}=y-(k-p)\end{align*}

Your job is to simplify and algebraically manipulate this equation to show that it is equivalent to \begin{align*}(x-h)^2=4p(y-k)\end{align*}. Start by squaring both sides of the equation, then expand all of the terms involving \begin{align*}y\end{align*}, \begin{align*}k\end{align*}, and \begin{align*}p\end{align*}, then simplify.

\begin{align*}\left(\sqrt{(x-h)^2+(y-(k+p) )^2}\right)^2 &= (y-(k-p) )^2 \\ (x-h)^2+(y-(k+p) )^2 &= (y-(k-p) )^2 \\ (x-h)^2+y^2-2py+p^2-2ky+2kp+k^2 &= y^2+2py+p^2-2ky-2kp+k^2 \\ (x-h)^2 &= 4py-4kp \\ (x-h)^2 &= 4p(y-k)\end{align*}

This is the equation of the parabola because the set of points on the parabola (which are the set of points that are the same distance from the focus as from the directrix), are the set of points that will satisfy this equation.

**Examples**

**Example 1**

Earlier, you were asked how the shape of a parabola changes as its focus and directrix move further apart.

As the focus and directrix move further apart, the parabola opens wider.

#### Example 2

Sketch the parabola \begin{align*}(x-2)^2=4(y-5) \end{align*} using the vertex, focus, and directrix.

The vertex is at (2, 5) and \begin{align*}p\end{align*} is 1. Plot the vertex and then find the focus and directrix. Finally, sketch in the parabola.

#### Example 3

Sketch the parabola \begin{align*}(x-2)^2=-4(y-5)\end{align*} using the vertex, focus, and directrix. What does it mean to have a negative value for \begin{align*}p\end{align*}?

A negative value for \begin{align*}p\end{align*} means that the directrix is above the focus and the parabola opens down. The vertex is in the same place, but the focus and directrix switch places.

#### Example 4

What is the equation of the parabola with focus (6, 1) and directrix \begin{align*}x=2\end{align*}?

This parabola opens to the right. The vertex is in the middle of the focus and directrix, so the vertex is at (4, 1). \begin{align*}p\end{align*} is 2. The equation is:

\begin{align*}(y-1)^2=8(x-4)\end{align*}

### Review

1. What is the relationship between a parabola and its focus and directrix?

For each equation below, state whether the parabola will open up, down, left, or right.

2. \begin{align*}(x-4)^2=8(y+1)\end{align*}

3. \begin{align*}(y+1)^2=-8(x-4)\end{align*}

4. \begin{align*}(x-4)^2=-8(y+1)\end{align*}

5. \begin{align*}(y+1)^2=8(x-4)\end{align*}

Sketch each parabola using its vertex, focus, and directrix.

6. \begin{align*}(x-4)^2=8(y+1)\end{align*}

7. \begin{align*} (y+1)^2=-8(x-4)\end{align*}

8. \begin{align*}(x-4)^2=-8(y+1)\end{align*}

9. \begin{align*}(y+1)^2=8(x-4)\end{align*}

10. What is the equation of the parabola with focus (4, 1) and directrix \begin{align*} y=-3\end{align*}?

11. What is the equation of the parabola with focus (3, 3) and directrix \begin{align*}y=-5\end{align*}?

12. What is the equation of the parabola with focus (2, 4) and directrix \begin{align*}x=4\end{align*}?

Consider the sideways parabola below.

13. What is the equation of the focus and the equation of the directrix in terms of \begin{align*}h\end{align*}, \begin{align*}k\end{align*}, and \begin{align*}p\end{align*}?

14. Consider a random point \begin{align*}(x, y)\end{align*} on the parabola. Find the distance from the point to the focus (see Example B for help).

15. Consider the same random point \begin{align*}(x, y)\end{align*} on the parabola as in #14. Find the perpendicular distance from the point to the directrix.

16. Use your answers to #14 and #15 to derive the equation for the sideways parabola: \begin{align*}(y-k)^2=4p(x-h)\end{align*}.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 10.3.