Lesson 7

What do you notice? What do you wonder?

The applet shows a parabola. In the applet, move point \(F\) (the focus) and the line (the directrix) and observe how the shape of the parabola changes.

1. What happens as the focus and directrix move farther apart?
2. Try to make the parabola open downward (that is, to look like a hill instead of a valley). What needs to be true for this to happen?
3. The vertex of the parabola is the lowest point on the curve if it opens upward, or the highest point if it opens downward. Where is the vertex located in relationship to the focus and the directrix?
4. Move the directrix to lie on the \(x\)-axis and move the focus to be on the point \((2,2)\). Plot a point \(P\), with coordinates \((6,5)\). It should lie on the parabola.
   1. What is the distance between point \(P\) and the directrix?
   2. What does this tell you about the distance between \(P\) and \(F\)?

The image shows a parabola with focus \((6,4)\) and directrix \(y=0\) (the \(x\)-axis).

1. The point \((11, 5)\) looks like it might be on the parabola. Determine if it really is on the parabola. Explain or show your reasoning.
2. The point \((14,10)\) looks like it might be on the parabola. Determine if it really is on the parabola. Explain or show your reasoning.
3. In general, how can you determine if a particular point \((x,y)\) is on the parabola?

The diagram shows several points that are the same distance from the point \((2,1)\) as they are from the line \(y=\text-3\)​​​. (Distance is measured from each point to the line ​​along a segment perpendicular to the line.) The set of all points that are the same distance from a given point and a given line form a parabola. The given point is called the parabola’s focus and the line is called its directrix.

We can use this definition to test if points are on a parabola. The image shows the parabola with focus \((2,3)\) and directrix \(y=1\). The point \((\text-2,6)\) appears to be on the parabola. Counting downwards, the distance between \((\text-2,6)\) and the directrix is 5 units.

Now use the Pythagorean Theorem to find the distance \(d\) between \((\text-2,6)\) and the focus, \((2,3)\). Imagine drawing a right triangle whose hypotenuse is the segment connecting \((\text-2,6)\) and \((2,3)\). The lengths of the triangle’s legs can be found by subtracting the corresponding coordinates of the points.

Use those lengths in the Pythagorean Theorem to get \((\text-2-2)^2+(6-3)^2=d^2\). Evaluate the left side of the equation to find that \(25=d^2\). The distance, then, is 5 units because 5 is the positive number that squares to make 25. Now we know the point \((\text-2,6)\) really is on the parabola, because it’s 5 units away from both the focus and the directrix.