Lesson 11.2: Parabolas

New Concept

Parabolas are U-shaped curves defined by their equations. You will learn to graph both vertical ($y = ax^2+…$) and horizontal ($x = ay^2+…$) types by analyzing their properties, and apply this to solve real-world problems.

What’s next

Next, you'll work through interactive examples for graphing parabolas. A series of practice cards will help you master the steps for each type.

Property

A parabola is all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.

Examples

• The path of a basketball shot through the air follows a parabolic arc due to gravity.
• A satellite dish is shaped like a parabola to gather incoming signals and reflect them to a single point, the focus, where the receiver is located.
• The reflector in a car's headlight has a parabolic shape to take the light from the bulb (at the focus) and project it forward in a strong, straight beam.

Explanation

Think of a parabola as a perfect U-shape. Every point on this curve is an equal distance away from a special point (the focus) and a special line (the directrix). This unique balance creates the parabola's signature curve.

Property

For a vertical parabola in general form $y = ax^2 + bx + c$ or standard form $y = a(x - h)^2 + k$:

• Orientation: Opens upward if $a > 0$; opens downward if $a < 0$.
• Axis of Symmetry: The vertical line $x = -\frac{b}{2a}$ or $x = h$.
• Vertex: The point $(h, k)$. For the general form, find it by substituting the x-value from the axis of symmetry into the equation.

To graph, determine the orientation, find the axis of symmetry and vertex, find the y-intercept (where $x=0$) and a symmetric point, find the x-intercepts (where $y=0$), and then draw the curve.

Examples

• For $y = x^2 - 6x + 5$, the parabola opens up since $a=1 > 0$. The axis of symmetry is $x = -\frac{-6}{2(1)} = 3$. The vertex is at $(3, -4)$.
• For $y = -2(x - 1)^2 + 8$, the parabola opens down since $a=-2 < 0$. The vertex is $(1, 8)$, and the axis of symmetry is the line $x=1$.
• To put $y = 3x^2 + 12x + 11$ in standard form, complete the square: $y = 3(x+2)^2 - 1$. This is an upward-opening parabola with its vertex at $(-2, -1)$.

Property

For a horizontal parabola in general form $x = ay^2 + by + c$ or standard form $x = a(y - k)^2 + h$:

• Orientation: Opens to the right if $a > 0$; opens to the left if $a < 0$.
• Axis of Symmetry: The horizontal line $y = -\frac{b}{2a}$ or $y = k$.
• Vertex: The point $(h, k)$. For the general form, find it by substituting the y-value from the axis of symmetry.

To graph, determine the orientation, find the axis of symmetry and vertex, find the x-intercept (where $y=0$), and find the y-intercepts (where $x=0$).

Examples

• For $x = y^2 + 4y + 3$, the parabola opens right since $a=1 > 0$. The axis of symmetry is $y = -\frac{4}{2(1)} = -2$. The vertex is at $(-1, -2)$.
• For $x = -3(y - 2)^2 + 5$, the parabola opens left since $a=-3 < 0$. The vertex is $(5, 2)$, and the axis of symmetry is the line $y=2$.
• To put $x = 2y^2 - 8y + 9$ in standard form, complete the square: $x = 2(y-2)^2 + 1$. This is a right-opening parabola with its vertex at $(1, 2)$.

Property

To find the equation of a parabolic arch, first set up a coordinate system. Place the vertex $(h, k)$ at the highest point of the arch. Use another known point on the arch, such as a base point $(x, y)$, to substitute into the standard form $y = a(x - h)^2 + k$ and solve for the value of $a$. Once $a$ is known, you have the complete equation for the arch.

Examples

• A parabolic arch is 40 ft wide and 10 ft high. With the vertex at $(20, 10)$ and a base at $(0, 0)$, we solve $0 = a(0-20)^2 + 10$ to get $a = -\frac{1}{40}$. The equation is $y = -\frac{1}{40}(x - 20)^2 + 10$.
• An arch is 100 ft wide and 25 ft high. With the vertex at $(50, 25)$ and base at $(0, 0)$, we solve $0 = a(0-50)^2 + 25$ to find $a = -\frac{1}{100}$. The equation is $y = -\frac{1}{100}(x - 50)^2 + 25$.
• A parabolic tunnel entrance is 30 ft wide and 20 ft high. With the vertex at $(15, 20)$ and base at $(0, 0)$, we solve $0 = a(0-15)^2 + 20$ to find $a = -\frac{20}{225} = -\frac{4}{45}$. The equation is $y = -\frac{4}{45}(x - 15)^2 + 20$.

Explanation

Many real-world arches, like those in bridges, are parabolic. By placing the arch on a coordinate grid, you can use the vertex and one other point to create a precise mathematical equation that models its shape.