Conic Sections: The Parabolas Lesson

In this lesson, we will learn about the geometric definition of a parabola. Before we get into that definition, let's quickly recap what we previously learned about parabolas.

Vertical Parabola

$$y = ax^2 + bx + c$$ Vertex Formula: $$h = -\frac{b}{2a}, k = c - \frac{b^2}{4a}$$ Vertex Form: $$y = a(x - h)^2 + k$$

• If a > 0, the parabola opens up
• If a < 0, the parabola opens down
• The vertex occurs at (h, k)
• Axis of symmetry: x = h

Horizontal Parabola

$$x = ay^2 + by + c$$ Vertex Formula: $$k = -\frac{b}{2a}, h = c - \frac{b^2}{4a}$$ Vertex Form: $$x = a(y - k)^2 + h$$

• If a > 0, the parabola opens right
• If a < 0, the parabola opens left
• The vertex occurs at (h, k)
• Axis of symmetry: y = k

Geometric Definition of a Parabola

A parabola is the set of all points in a plane equidistant from a fixed point, known as the focus, and a fixed line, known as the directrix. The axis of symmetry will pass through the focus and is perpendicular to the directrix. The point of intersection of the parabola with its axis of symmetry is known as its vertex. Additionally, the vertex is the midpoint of the line segment that joins the focus and directrix on the axis of symmetry. From the image above, we can see that for any point P(x, y) on the parabola, the distance (d₁) to the focus F(0, p) is equal to the distance (d₂) to the directrix D(x, -p). Using this definition, we can find the equation of a parabola. $$d(P, F) = d(P, D)$$ Let's start by finding the distance between points P(x, y) and F(0, p).
Using the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$d(P, F) = \sqrt{(x - 0)^2 + (y - p)^2}=\sqrt{x^2 + y^2 - 2py + p^2}$$ Now, let's find the distance between points P(x, y) and D(x, -p).
Using the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$d(P, D) = \sqrt{(x - x)^2 + (y - (-p))^2} = \sqrt{(y + p)^2} = \sqrt{y^2 + 2py + p^2}$$ Set the two equal and simplify: $$d(P, F) = d(P, D)$$ $$\sqrt{x^2 + y^2 - 2py + p^2} = \sqrt{y^2 + 2py + p^2}$$ $$\left(\sqrt{x^2 + y^2 - 2py + p^2}\right)^2 = \left(\sqrt{y^2 + 2py + p^2}\right)^2$$ $$x^2 + y^2 - 2py + p^2 = y^2 + 2py + p^2$$ $$\require{cancel}x^2 + \cancel{y^2} - 2py + \cancel{p^2} = \cancel{y^2} + 2py + \cancel{p^2}$$ $$x^2 - 2py = 2py$$ $$x^2 = 4py$$

Vertical Parabola with its Vertex at the Origin

The parabola with focus (0, p) and directrix y = -p has the equation: $$x^2 = 4py$$

• Axis of symmetry: x = 0
• The parabola opens up if p > 0
• The parabola opens down if p < 0

Horizontal Parabola with its Vertex at the Origin

Using a similar process, we can derive an equation for a horizontal parabola using the geometric definition given earlier. The parabola with focus (p, 0) and directrix x = -p has the equation: $$y^2 = 4px$$

• Axis of symmetry: y = 0
• The parabola opens right if p > 0
• The parabola opens left if p < 0

Let's look at some examples.
Example #1: Find the focus and directrix. $$y = -2x^2$$ Since this is a vertical parabola with its vertex at the origin, let's match the form given above. $$x^2 = 4py$$ $$y = -2x^2$$ Multiply both sides by -1/2: $$-\frac{1}{2}y = x^2$$ Switch sides: $$x^2 = -\frac{1}{2}y$$ Now we can compare: $$x^2 = 4py$$ $$x^2 = -\frac{1}{2}y$$ Find p: $$4p = -\frac{1}{2}$$ $$p = -\frac{1}{2} \cdot \frac{1}{4} = -\frac{1}{8}$$ The best way to think about p is as a directed distance from the vertex to the focus. In this case, we know that our vertex is at the origin, and since p is negative, the parabola opens down. So we would travel down 1/8 units from the origin to get to the focus. $$\text{Focus:} \, \left(0, -\frac{1}{8}\right)$$ The directrix is given as y = -p: $$y = -\left(-\frac{1}{8}\right) = \frac{1}{8}$$ $$\text{Directrix:} \, y = \frac{1}{8}$$ Desmos Link for More Detail Example #2: Find the focus and directrix. $$x = \frac{1}{6}y^2$$ Since this is a horizontal parabola with its vertex at the origin, let's match the form given above. $$y^2 = 4px$$ $$x = \frac{1}{6}y^2$$ Multiply both sides by 6: $$6x = y^2$$ Switch sides: $$y^2 = 6x$$ Now we can compare: $$y^2 = 4px$$ $$y^2 = 6x$$ Find p: $$4p = 6$$ $$p = \frac{6}{4} = \frac{3}{2}$$ The best way to think about p is as a directed distance from the vertex to the focus. In this case, we know that our vertex is at the origin, and since p is positive, the parabola opens to the right. So we would travel to the right by 3/2 units from the origin to get to the focus. $$\text{Focus:} \, \left(\frac{3}{2}, 0\right)$$ The directrix is given as x = -p: $$x = -\frac{3}{2}$$ $$\text{Directrix:} \, x = -\frac{3}{2}$$ Desmos Link for More Detail Example #3: Write an equation for the parabola, give the focus and directrix.

• Vertical parabola
• Vertex at the origin: (0, 0)
• Given point: (5, 5)

Since this is a vertical parabola with its vertex at the origin, we will reference our formula above. $$x^2 = 4py$$ Plug in the given point: (5, 5) $$(5)^2 = 4p(5)$$ Solve for p: $$25 = 20p$$ $$p = \frac{25}{20} = \frac{5}{4}$$ Plug in for p: $$x^2 = 4\left(\frac{5}{4}\right)y$$ $$x^2 = 5y$$ $$\text{Focus:} \, \left(0, \frac{5}{4}\right)$$ $$\text{Directrix:} \, y = -\frac{5}{4}$$ Desmos Link for More Detail

Translations of Parabolas

In many cases, the parabola will not have its vertex at the origin. When this situation occurs, we can replace x with (x - h) and y with (y - k). This gives us the following:
Vertical Parabola: $$(x - h)^2 = 4p(y - k)$$ Horizontal Parabola: $$(y - k)^2 = 4p(x - h)$$ The focus is |p| units from the vertex.
Let's look at an example.
Example #4: Write the equation of the parabola and find the directrix. $$\text{Vertex:} \, (-9, -2)$$ $$\text{Focus:} \, \left(-9, -\frac{5}{4}\right)$$ First, we want to determine if this is a vertical or horizontal parabola. Since the x-coordinate of the vertex and the focus are the same, this means we will have a vertical parabola. $$(x - h)^2 = 4p(y - k)$$ Plug in for h and k: $$h = -9, k = -2$$ $$(x - (-9))^2 = 4p(y - (-2))$$ $$(x + 9)^2 = 4p(y + 2)$$ Recall that the focus is |p| units from the vertex. It might help to make a rough sketch. $$-\frac{5}{4} - \left(-2\right) = -\frac{5}{4} - \left(-\frac{8}{4}\right) = -\frac{5}{4} + \frac{8}{4} = \frac{3}{4}$$ To go from the vertex V(-9, -2) to the focus F(-9, -5/4), we need to travel up by 3/4 units. As we stated earlier, the best way to think about p is as a directed distance from the vertex to the focus. This tells us that p is 3/4. $$p = \frac{3}{4}$$ Plug in for p: $$(x + 9)^2 = 4p(y + 2)$$ $$(x + 9)^2 = 4 \cdot \frac{3}{4}(y + 2)$$ Our equation for the parabola: $$(x + 9)^2 = 3(y + 2)$$ To find the directrix, recall that the distance from any point to the focus is the same as the distance from that same point to the directrix. So from the vertex, we traveled up by 3/4 units to get to the focus. This means we would travel down by 3/4 units to get to the directrix. The x-value would be the same, so take the y-value from the vertex, which is -2, and subtract away 3/4. $$-2 - \frac{3}{4} = -\frac{8}{4} - \frac{3}{4} =-\frac{11}{4}$$ $$\text{Directrix:} \, y=-\frac{11}{4}$$ Desmos Link for More Detail Using a table to organize the formulas can be helpful. This will allow us to quickly find the focus and directrix once we have the vertex and the p-value. Let's look at an example.
Example #5: Find the vertex, focus, and directrix. $$x^2 - 2x + y + 3 = 0$$ We see that we have x² here which means we are dealing with a vertical parabola. $$(x - h)^2 = 4p(y - k)$$ To achieve the form above, subtract y away from each side and subtract 3 away from each side. $$x^2 - 2x = -y - 3$$ Complete the square on the left. $$x^2 - 2x + 1 = -y - 3 + 1$$ $$(x - 1)^2 = -y - 2$$ Factor out the -1 from the right side. $$(x - 1)^2 = -(y + 2)$$ We will grab our values for h and k. $$(x - h)^2 = 4p(y - k)$$ $$(x - 1)^2 = -1(y - (-2))$$ $$h = 1, k = -2$$ $$\text{Vertex: } \, (1, -2)$$ Now, let's get our p-value. $$4p = -1$$ $$p = -\frac{1}{4}$$ To get the focus, use the formula: $$\text{Focus:} \, (h, k + p)$$ $$k + p = -2 + -\frac{1}{4} = -\frac{8}{4} + -\frac{1}{4} = -\frac{9}{4}$$ $$\text{Focus:} \, \left(1, -\frac{9}{4}\right)$$ $$\text{Directrix:} \, y = k - p$$ $$k - p = -2 - \left(-\frac{1}{4}\right) = -\frac{8}{4} + \frac{1}{4} = -\frac{7}{4}$$ $$\text{Directrix:} \, y = -\frac{7}{4}$$ Desmos Link for More Detail

Latus Rectum

The latus rectum of a parabola is a line segment with endpoints on the parabola that passes through the focus and is parallel to its directrix. To find the length of the latus rectum, we can start with a horizontal parabola with its vertex at the origin. $$y^2 = 4px$$ From the definition, the endpoints are on the parabola and it passes through the focus. This means the x-coordinate of each endpoint will be the same as the x-coordinate of the focus. $$\text{Focus:} \, (p, 0)$$ $$\text{Endpoint 1:} \, (p, y_1)$$ $$\text{Endpoint 2:} \, (p, y_2)$$ Since we know the endpoints are on the parabola and have an x-coordinate of p in each case, we can just plug in a p for x and solve for y. $$y^2 = 4px$$ $$y^2 = 4p(p)$$ $$y^2 = 4p^2$$ Solve for y: $$y = \pm \sqrt{4p^2}$$ $$y = \pm 2|p|$$ $$y = \pm 2p$$ We can now update our endpoints: $$\text{Endpoint 1:} \, (p, 2p)$$ $$\text{Endpoint 2:} \, (p, -2p)$$ So the distance from (p, 2p) to (p, -2p) is found as: $$2p - (-2p) = 2p + 2p = 4p$$

• The focus is the midpoint of the latus rectum
  • Recall that the axis of symmetry passes through the focus
  • The distance from the focus to each endpoint will be the same
• The length of the latus rectum is |4p|
• For a vertical parabola, we find the endpoints as:
  • Travel left from the focus by |2p|
  • Travel right from the focus by |2p|
• For a horizontal parabola, we find the endpoints as:
  • Travel down from the focus by |2p|
  • Travel up from the focus by |2p|

Let's look at an example.
Example #6: Find the two endpoints of the latus rectum. $$(y - 3)^2 = 8(x + 1)$$ Let's grab our vertex: (h, k). $$(y - k)^2 = 4p(x - h)$$ $$(y - 3)^2 = 8(x - (-1))$$ $$h = -1, k = 3$$ Let's grab our p-value. $$4p = 8$$ $$p = \frac{8}{4} = 2$$ $$\text{Focus:} \, (h + p, k)$$ $$h + p = -1 + 2 = 1$$ $$\text{Focus:} \, (1, 3)$$ The length of the latus rectum is |4p|. $$|4p| = |4 \cdot 2| = |8| = 8$$ The graph will have horizontal symmetry, so the latus rectum will extend 4 units up and 4 units down from the focus. This can also be found using the formula above. We will travel up by |2p| units and down by |2p| units from the focus. $$|2p| = |2 \cdot 2| = 4$$ Endpoints of the latus rectum: $$(1, 3 + 4) = (1, 7)$$ $$(1, 3 - 4) = (1, -1)$$ Desmos Link for More Detail

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