Equation of a Parabola

The equation of a parabola represents a U-shaped curve formed by the graph of a quadratic function. It’s widely used in mathematics, physics, and engineering to model real-life situations such as projectile motion, satellite dishes, and optics. Understanding how to derive, graph, and transform the equation of a parabola is crucial for both high school and competitive exams.

1.0What is a Parabola?

A parabola is a set of points that are equidistant from a fixed point (focus) and a fixed line (directrix). It opens either up, down, left, or right, depending on the orientation of the axis of symmetry.

2.0Equation of a Parabola in Standard Form

The standard form of a parabola depends on the axis of symmetry:

1. Vertical Parabola (opens up/down): $(x-h{)}^2=4a(y-k)$
2. Horizontal Parabola (opens left/right): $(y-k{)}^2=4a(x-h)$

Where:

• (h, k) is the vertex,
• aa is the distance from the vertex to the focus,
• The directrix is a fixed line located opposite the focus.

3.0Equation of a Parabola in Vertex Form

For a vertical parabola: $y=a(x-h{)}^2+k$

Here:

• aa controls the direction and width of the parabola,
• (h, k) is the vertex.

Determining the Equation of a Parabola:

To determine the equation, you need:

• The vertex of the parabola,
• A point on the curve (or focus and directrix),
• The direction it opens.

4.0Parametric Equation of a Parabola

For the parabola $(x-h{)}^2=4a(y-k)$, the parametric equations are: $x=h+2at,y=k_a{t}^2$

Where t is the parameter.

5.0Graphing Parabola Equations

Graphing parabola equations involves:

1. Identifying the vertex.
2. Determining the direction the parabola opens.
3. Plotting the focus and directrix.
4. Drawing the symmetric curve.

Use a graphing calculator or plotting software for accuracy, especially when the parabola is rotated or translated.

Also Explore: Graphical Method Linear Programming

6.0Parabola Equation Examples

Example 1: Find the equation of a parabola with vertex at (0, 0) and focus at (0, 3).

Solution:

• Since it opens upward, use: $(x-h{)}^2=4a(y-k)$
• a = 3, h = 0, k = 0

$\Rightarrow x^2=12y$

Example 2: Find the equation of the directrix of the parabola $y=2(x+1{)}^2+3$

Solution:

This is in vertex form, vertex is (1, 3), a = 2

• Directrix: $y=k-\frac{1}{4a}=3-\frac{1}{8}=\frac{23}{8}$
  

Example 3: Find the equation of the parabola whose focus is at (3, 0) and directrix is the line x = -3. Also find its vertex.

Solution:

• The vertex lies midway between focus and directrix: $V=(\frac{3+(-3)}{2},0)=(0,0)$
• Distance from vertex to focus = 3
• So, a = 3, opening rightwards.
• Standard form: $y^2=4ax\Rightarrow y^2=12x$

Answer: $y^2=12x$; Vertex: (0, 0)

Example 4: Find the equation of the tangent to the parabola $y^2=4x$ that makes an angle of $45^0$ with the x-axis.

Solution:

• Slope of line = $m=tan(45^0)=1$
• Equation of tangent in slope form:

$y=mx+\frac{a}{m}$, where a = 1

$\Rightarrow y=x+\frac{1}{1}=x+1$

Answer: y = x + 1

Example 5: A chord of the parabola $y^2=4ax$ passes through the focus and is perpendicular to the axis. Find its length.

Solution:

• Focus: (a, 0)
• Line perpendicular to x-axis through focus: x = a
• Sub into parabola: $y^2=4a^2\Rightarrow y=\pm 2a$
• Points: (a, 2a), (a, -2a)
• Length $=2\times 2a=4a$

Answer: 4a

Example 6: Find the coordinates of the foot of perpendicular from the focus of the parabola $y^2=8x$ on the line y = x + 2.

Solution:

• Focus of $y^2=8x$: (2, 0)
• Perpendicular from (2, 0) to y = x + 2:

Slope of given line = 1 → slope of perpendicular = -1

• Equation of perpendicular:

Passes through (2, 0), slope -1 $\Rightarrow y-0=-1(x-2)\Rightarrow y=-x+2$

• Solve with y = x + 2 and y = -x + 2 $\Rightarrow x+2=-x+2\Rightarrow 2x=0\Rightarrow x=0,y=2$

Answer: (0, 2)

Example 7: Find the locus of a point which moves such that its distance from the point (0, 3) is equal to its distance from the line y = -3.

Solution:
Let point be (x, y)

• Distance from (0, 3): $\sqrt{x^2+(y-3{)}^2}$
• Distance from line $y=-3y=-3:|y+3||y+3|$

Equating: $\sqrt{x^2+(y-3{)}^2}=|y+3|$

Squaring:

$x^2+(y-3{)}^2=(y+3{)}^2$

$\Rightarrow x^2+y^2-6y+9=y^2+6y+9$

$\Rightarrow x^2=12y$

Answer: $x^2=12y$ a vertical parabola

7.0Practice Questions on Equation of Parabola

1. Convert $x=-2y^2+4y-1$ to standard form and identify vertex and axis.
2. Find the parametric form of the parabola $x^2=8y$.
3. Graph the parabola $y=-3(x+1{)}^2+2$.
4. Determine the equation of the directrix for $x^2=-16y$.