Parabola

A parabola is a curve you’ve already seen many times before. It’s the shape of the graph of a quadratic function.

However, when we talk about parabolas in the context of conic sections, we usually mean an equation in the form y^2=4ax. It’s a parabola that opens to the right, and the vertex is at the origin. Basically: a sideways parabola.

Equation of a parabola

As mentioned before, the equation of a right-opening parabola is:

y^2 = 4ax

a is the distance from the vertex to the focus (the point where the parabola is “pointiest”).

Examples

Find the equation of the parabola with focus at (2, 0)

The distance from the vertex to the focus is 2, so a = 2.
y^2 = 4 \cdot 2 \cdot x = 8x.

Find the equation of the parabola with focus at (5, 0)

The distance from the vertex to the focus is 5, so a = 5.
y^2 = 4 \cdot 5 \cdot x = 20x.

Find the equation of the parabola with focus at (0, 0)

The distance from the vertex to the focus is 0, so a = 0.
y^2 = 4 \cdot 0 \cdot x = 0.

Find the focus of the parabola with equation y^2 = 12x

4a = 12 \Rightarrow a = 3.
The focus is at (3, 0).

Find the focus of the parabola with equation y^2 = 32x

4a = 32 \Rightarrow a = 8.
The focus is at (8, 0).

Find the focus of the parabola with equation y^2 = 0

4a = 0 \Rightarrow a = 0.
The focus is at (0, 0).

flashcards

Question	Answer
What is the general form of a parabola in conic sections?	y^2 = 4ax
What does a represent in the parabola equation y^2 = 4ax?	The distance from the vertex to the focus.
What is the focus of a parabola?	The point where the parabola is “pointiest”, located at (a, 0).
Find the equation of the parabola with focus at (2, 0).	a = 2, so y^2 = 4 \cdot 2 \cdot x = 8x
Find the equation of the parabola with focus at (5, 0).	a = 5, so y^2 = 4 \cdot 5 \cdot x = 20x
Find the equation of the parabola with focus at (0, 0).	a = 0, so y^2 = 4 \cdot 0 \cdot x = 0
Find the focus of the parabola with equation y^2 = 12x.	4a = 12 \Rightarrow a = 3, focus is at (3, 0)
Find the focus of the parabola with equation y^2 = 32x.	4a = 32 \Rightarrow a = 8, focus is at (8, 0)
Find the focus of the parabola with equation y^2 = 0.	4a = 0 \Rightarrow a = 0, focus is at (0, 0)
What is the vertex of the parabola y^2 = 4ax?	The origin (0, 0)
Which direction does the parabola y^2 = 4ax open?	It opens to the right.
True or False: A parabola is the shape of the graph of a quadratic function.	True