Keoni and Sasha use the Pythagorean Theorem and the definition of a parabola to derive an equation of parabola with a p-value of 5 and a vertex at (9, 13).
Students’ Conceptual Challenges
At [0:48], Keoni is unsure whether to draw a segment from the general point to the vertex or to the focus.
➤ Sasha and Keoni resolve the issue by reminding themselves of the definition of a parabola. They recall how they can use the definition to represent the lengths of the sides of a right triangle.
For use in a classroom, pause the video and ask these questions:
1. [Pause the video at 1:54]. What is the distance y? How would you represent it geometrically on the coordinate grid?
2. [Pause the video at 3:46]. What are the coordinates of the third vertex of the triangle?
Invite students to engage in productive disagreement:
As Sasha and Keoni are completing their algebraic manipulations, Sasha cancels the 20 [7:34]. But then it shows up in the final version of the equation. What happened? Did she make a mistake?
In the graph below, the base parabola with p =5, shown in black, is represented by the equation y = x2/20. If the base parabola is translated to the right 9 units, it will be the parabola shown in light blue, which is represented by the equation y = (x–9)2/20. If this parabola is translated up 13 units, it will be the parabola shown in dark blue, which is represented by the equation y = (x–9)2/20 + 13.
1. Use the equations to prove that each vertex is on a particular parabola; what do you notice?
2. Use the equations to find a “special point” (a point aligned horizontally with the focus) for each parabola; what do you notice?