Conic Sections Reference

Parabolas

A parabola can be studied as a quadratic graph or as a conic section. The quadratic view emphasizes vertex, intercepts, and transformations; the conic view explains focus, directrix, focal width, and the latus rectum.

Solved Examples

Jump directly to a worked example by parabola type.

Quadratic Summary Table

Form What It Shows Use It For
$y=ax^2+bx+c$ Standard form; y-intercept is $c$. Axis formula, discriminant, quick y-intercept.
$y=a(x-h)^2+k$ Vertex form; vertex is $(h,k)$. Transformations, graphing, maximum or minimum.
$y=a(x-r_1)(x-r_2)$ Factored form; zeros are $r_1$ and $r_2$. X-intercepts and sign behavior.
$a>0$ Opens upward. Vertex is a minimum.
$a<0$ Opens downward. Vertex is a maximum.

Quadratic Formulas

Axis of Symmetry
$$x=-\frac{b}{2a}$$
Vertex from Standard Form
$$h=-\frac{b}{2a},\quad k=f(h)$$
Vertex Form
$$y=a(x-h)^2+k$$
Conic Connection
$$y=a(x-h)^2+k\quad\Longleftrightarrow\quad a=\frac1{4p}$$
The value of $|a|$ controls how narrow or wide a vertical parabola looks. In conic language, that same width is measured by $p$ and the focal width $|4p|$.

Quadratic Classic Examples

Vertex and Axis

Find the vertex and axis of symmetry for $f(x)=2x^2-8x+3$.

Solution Steps
  1. Use the axis formula with $a=2$ and $b=-8$.
  2. Evaluate the axis coordinate.
  3. Substitute the axis value into the function.
  4. Evaluate the quadratic at the axis value.
  5. State the vertex.
  6. State the axis of symmetry.
  7. State the final vertex and axis together.
$$h=-\frac{-8}{2(2)}$$ $$=2$$ $$k=f(2)$$ $$=2(2)^2-8(2)+3$$ $$=-5$$ $$\text{vertex: }(2,-5)$$ $$\text{axis: }x=2$$

Read Transformations

Describe $g(x)=-3(x+1)^2+4$.

Solution Steps
  1. Rewrite the expression to expose vertex form.
  2. Read the vertex.
  3. Use the negative coefficient to identify downward opening.
  4. Read the vertical stretch factor.
  5. State the maximum value.
$$g(x)=-3(x-(-1))^2+4$$ $$\text{vertex: }(-1,4)$$ $$\text{opens downward}$$ $$\text{vertical stretch by }3$$ $$\text{maximum value: }4$$

Conic Summary Table

Form Opens Focus and Directrix
$(x-h)^2=4p(y-k)$ Up if $p>0$, down if $p<0$. Focus $(h,k+p)$, directrix $y=k-p$.
$(y-k)^2=4p(x-h)$ Right if $p>0$, left if $p<0$. Focus $(h+p,k)$, directrix $x=h-p$.
Vertex Turning point. $(h,k)$
Axis of symmetry Line through vertex and focus. $x=h$ for vertical parabolas; $y=k$ for horizontal parabolas.
Latus rectum Segment through the focus perpendicular to the axis. Length $|4p|$.

Conic Formulas

Vertical Axis
$$(x-h)^2=4p(y-k)$$ $$\text{focus: }(h,k+p)$$ $$\text{directrix: }y=k-p$$
Horizontal Axis
$$(y-k)^2=4p(x-h)$$ $$\text{focus: }(h+p,k)$$ $$\text{directrix: }x=h-p$$
Focal Width
$$\text{latus rectum length}=|4p|$$
Parameter
$$p=\text{directed distance from vertex to focus}$$
The squared variable tells us the direction: if $x$ is squared, the parabola opens up or down; if $y$ is squared, it opens left or right.

Conic Classic Examples

Find Focus and Directrix

Find the vertex, focus, directrix, axis, and focal width of $(x-2)^2=12(y+1)$.

Solution Steps
  1. Match the equation to the vertical standard form.
  2. Read $h$, $k$, and $4p$.
  3. Solve for $p$.
  4. State the vertex.
  5. Move $p$ units to find the focus.
  6. State the directrix.
  7. State the axis.
  8. State the focal width.
$$(x-2)^2=12(y-(-1))$$ $$h=2,\quad k=-1,\quad 4p=12$$ $$p=3$$ $$\text{vertex: }(2,-1)$$ $$\text{focus: }(2,2)$$ $$\text{directrix: }y=-4$$ $$\text{axis: }x=2$$ $$\text{focal width: }12$$

Write an Equation

A parabola has vertex $(1,-2)$ and focus $(5,-2)$. Write its equation.

Solution Steps
  1. Read $h$, $k$, and the directed distance $p$ from the vertex to the focus.
  2. Choose the horizontal standard form.
  3. Substitute the values into the form.
  4. State the equation.
$$h=1,\quad k=-2,\quad p=4$$ $$(y-k)^2=4p(x-h)$$ $$(y+2)^2=4(4)(x-1)$$ $$(y+2)^2=16(x-1)$$

Latus Rectum Summary Table

Parabola Type Latus Rectum Line Endpoints
$(x-h)^2=4p(y-k)$ $y=k+p$ $(h-2p,k+p)$ and $(h+2p,k+p)$
$(y-k)^2=4p(x-h)$ $x=h+p$ $(h+p,k-2p)$ and $(h+p,k+2p)$

Latus Rectum Formulas

Vertical Parabola
$$\text{line: }y=k+p$$ $$(h\pm2p,k+p)$$
Horizontal Parabola
$$\text{line: }x=h+p$$ $$(h+p,k\pm2p)$$
Length
$$|4p|$$
Half-Length
$$|2p|$$
The latus rectum is sometimes misspelled as "lattice rectum." The standard term is latus rectum.

Latus Rectum Classic Examples

Vertical Latus Rectum

Find the latus rectum line and endpoints for $(x-2)^2=12(y+1)$.

Solution Steps
  1. Read $h$, $k$, and $p$ from standard form.
  2. Use the horizontal latus rectum line formula.
  3. Evaluate the line.
  4. Substitute into the first endpoint formula.
  5. Substitute into the second endpoint formula.
  6. State the latus rectum length.
$$h=2,\quad k=-1,\quad p=3$$ $$\text{line: }y=k+p$$ $$y=2$$ $$(h-2p,k+p)=(2-6,2)=(-4,2)$$ $$(h+2p,k+p)=(2+6,2)=(8,2)$$ $$\text{length: }12$$

Horizontal Latus Rectum

Find the latus rectum line and endpoints for $(y+2)^2=16(x-1)$.

Solution Steps
  1. Read $h$, $k$, and $4p$ from standard form.
  2. Solve for $p$.
  3. Use the vertical latus rectum line formula.
  4. Evaluate the line.
  5. Substitute into the first endpoint formula.
  6. Substitute into the second endpoint formula.
  7. State the latus rectum length.
$$h=1,\quad k=-2,\quad 4p=16$$ $$p=4$$ $$\text{line: }x=h+p$$ $$x=5$$ $$(h+p,k-2p)=(5,-2-8)=(5,-10)$$ $$(h+p,k+2p)=(5,-2+8)=(5,6)$$ $$\text{length: }16$$

Parabola Checklist

Step Question
1 Is the parabola being treated as a function graph or a conic section?
2 Which variable is squared?
3 What are $h$, $k$, and $p$?
4 Where are the vertex, focus, directrix, and axis?
5 What is the latus rectum line and what are its endpoints?
6 Does the graph need intercepts, domain, range, or maximum/minimum value?