Math101 · Independent Ontario learning library
Math101learn.math101.caParabola Equation
A parabola is the set of points equidistant from a focus and a directrix. Standard form $(x-h)^2=4p(y-k)$ opens vertically, while $(y-k)^2=4p(x-h)$ opens horizontally.
Focus-directrix equations explain reflectors, satellite dishes, headlights, and projectile graph shapes while linking analytic and synthetic geometry.
Intuition and core definition
A parabola is the set of points equidistant from a focus and a directrix. Standard form $(x-h)^2=4p(y-k)$ opens vertically, while $(y-k)^2=4p(x-h)$ opens horizontally. The vertex is $(h,k)$ and signed $p$ gives focus direction and distance.
Notation, language, and conditions
For vertical form, focus is $(h,k+p)$ and directrix $y=k-p$; $p>0$ opens up and $p<0$ down. For horizontal form, focus is $(h+p,k)$ and directrix $x=h-p$. The coefficient is $4p$, not $p$.
Why this idea matters
A parabola is equidistant from a focus and directrix, and its standard equation reveals vertex, orientation, and focal parameter.
A dependable method
- Identify which variable is squared to determine axis orientation.
- Read vertex from shifted coordinates.
- Set the coefficient of the unsquared displacement equal to $4p$.
- Use signed $p$ to locate focus and directrix.
- Check that the vertex is midway between focus and directrix and test a point.
Worked example
Representations and interpretation
The focus-directrix construction compares point-to-point and perpendicular point-to-line distances. Axis symmetry runs through focus and vertex; standard form compresses these geometric distances into an equation.
Reasoning about variations
Quadratic function form $y=a(x-h)^2+k$ can be rewritten $(x-h)^2=(1/a)(y-k)$, so $4p=1/a$. A large $|a|$ corresponds to small $|p|$ and a narrower graph.
Common mistakes
How to check your work
- Confirm equal vertex-to-focus and vertex-to-directrix distances.
- Substitute the vertex into the equation.
- Check focus direction matches the sign of $p$.
Practice
- For $x^2=20y$, find the focus.
- Find the directrix of $(y-2)^2=8(x+1)$.
- Which direction does $(x-4)^2=-16(y+3)$ open?
Answers and brief solutions
Show answers
- $(0,5)$ $4p=20$, so $p=5$ and focus is five units above the vertex.
- $x=-3$ Vertex $(-1,2)$ and $p=2$, so $x=h-p=-3$.
- Down $p=-4$ for a vertical parabola.
Synthesis and transfer
A parabolic reflector sends parallel incoming rays toward its focus; reading the focal distance from the equation determines where a receiver should be placed.
For $(y-2)^2=12(x+1)$, comparison with $(y-k)^2=4p(x-h)$ gives vertex $(-1,2)$ and $p=3$. The focus is $(2,2)$ and directrix is $x=-4$, so the parabola opens right. Testing the vertex shows equal distance $3$ to the focus and directrix, and another point can verify the locus property. A negative $p$ would reverse the opening direction without changing the focal distance magnitude. The axis passes through the vertex and focus, which helps distinguish this sideways form from a function written as $y=$ an expression of $x$.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
For $x^2=20y$, find the focus.
- $4p=20$, so $p=5$ and focus is five units above the vertex.
End of lesson
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