The intersection of a plane parallel to the edge of a nappe of a cone is called a parabola.
The red dashed lines are parallel, one is along the edge of the cone and the other is in the plane. The red curve is the intersection of the plane and the cone. (Note: the other nappe would be inverted tip to tip with this nappe and the plane is parallel with its edge as well but lies outside that nappe. In other words, this plane intersects only one nappe of the cone.)
Along the plane the intersection looks like the graph above. The dashed line in the plane is the parabola's axis of symmetry and the vertex is the point on the parabola that intersects its axis of symmetry. Each point on the parabola has a mirror image across the axis of symmetry.
Each parabola has an external line called a directrix which is perpendicular to the parabola's axis of symmetry and at a distance c from its vertex. The parabola also has a focus f along the axis of symmetry the same distance c from it vertex but opposite that of the directrix. The focus is always inside the parabola . The green lines shown above have equal length. The definition of a parabola is all those points whose perpendicular distances to the directrix equals their distances to the focus.
The equation for a parabola whose vertex is at the origin is: 


The gen
Interchanging x and y in this equation causes the parabola opens along the X axis like this:
If the vertex is translated to a point (h,k) the equation becomes:
Examples:
1) Plot the parabola: x^{2} = 12y Comparing to the general equation x^{2} = 4cy with vertex at the origin, and the axis of symmetry is the Y axis. 4c must equal 12. So c = 12/4 = 3, and since the axis of symmetry is the Y axis, the focal point is on the Y axis at y = 3 and the directrix is at y = c = 3. We get the shape by using symmetric points. Let's use x = 4 and x = 4, doing so we get (4)^{2} = (4)^{2} = 12y 16 = 12 y y = 16/12 = 4/3 Now we plot these points and sketch in the parabola. Sketch in the directrix, label the focus and a point on the parabola and its mirror image. 


2) Plot the parabola 3x^{2} = 24y First we need it in the correct form, x^{2} must have the coefficient 1. So, we divide both sides by 3 and get: x^{2} = 8y As in example 1, we have a parabola with vertex at the origin, and c = 2 (8 = 4c > 8/4 = c = 2) The axis of symmetry is the Y axis, so the focal point is at y=2 and the directrix is at y=2. Using x = 8 and 8, we get y = 8^{2}/8 = 8 for our symmetric points (8,8), (8,8)



3) Plot the parabola x^{2} = 6y Notice the negative coefficient for y. The short of it is, this parabola is the mirror image across the X axis of the parabola x^{2} = 6y. Here's a table of values to plot.
The plot of x^{2} = 6y is shown in red. Notice that it is the mirror image (across the X axis) of the parabola x^{2} = 6 shown in light blue. From the equation for a parabola with vertex at the origin we have 4c = 6, that is c = 6/4 = 3/2. So the focus is at (0, 3/2). Remember x = 0 for the vertex since the Y axis (x=0) is the axis of symmetry. The directrix is 3/2 from the vertex, along the Y axis, the directrix is the line y = 3/2. 





5) Plot (x + 3)^{2} = 16(y  2) This equation is almost in the general form: (x  h)^{2} = 4c ( y  k) first the left side: (x + 3) = (x  (3) ) so h = 3 no change required on the right hand side, k = 2 So the vertex is at (h, k) = (3, 2) 4c = 16, so c = 4 Since c is positive and x is the squared variable, this parabola opens up. x = 0 is a convenient point to take for the x coordinates of the symmetric points (this also gives the y intercept.) (0 + 3)^{2} = 16(y  2) 9 = 16(y  2) 9 = 16y  32 41 = 16y 41/16 = y So, our points of symmetry are (0, 2.56), (6, 2.56) (why x=6?) The axis of symmetry is the x value of the vertex which is 3. 0 is 3 units to the right, so the other x must be 3 units to the left of this axis, so 3 3 = 6. The directrix is horizontal at the distance c=4 from the vertex down. The vertex is at (3, 2) so 4 units down from this point is 2  4 = 2. 


6) Write the equation for this parabola.

The vertex is the origin and it opens along the positive Y axis. The equation we want is x^{2} = 4cy Two points are shown on the graph, (2,2) and (2,2). Using the first one and substituting into this equation we get: (2)^{2} = 4c(2) 4 = 8c 4/8 = c c = 1/2 So our equation is x^{2} = 4(1/2)y x^{2} = 2y Check the other point: (2)2 ? 2(2) yes, 4 = 4 ✓ 

7) Write the equation for this parabola.

This is a parabola whose vertex is not at the origin and it opens along the negative Y axis, so its equation is in the form: (x  h)^{2} = 4c(y  k) The vertex is at (2,2) which is (h,k). So far we have with substitution: (x  2)^{2} = 4c(y  (2)) = 4c(y + 2) Now we need c. The focal point is 'c' units from the vertex. From the graph f is 2 units away from the vertex, so c = 2. The equation is now (x  2)^{2} = 4(2)(y + 2) (x  2)^{2} = 8(y + 2)
Check: (2,4) ((2)  2)^{2} ? 8((4) + 2) (4)^{2} ? 8(2) 16 = 16 ✓ Note: c could have been found substituting the point (2,4) into the intermediate equation (x  2)^{2} = 4c(y + 2), then solve for c. The approach taken was more direct since the focal point was shown. 

8) Write the equation for this parabola.

This parabola opens along the positive X axis and its vertex is the origin. The general equation to use is: y^{2} = 4cx Any of the shown points can be used to find c, except the origin (in this case upon substitution, 0 = 0 ☺) Using (8,4) we have: (4)^{2} = 4c(8) 16 = 32c 16/32 = c c = 1/2 So, our equation becomes: y^{2} = 4 (1/2) x y^{2} = 2x
check: let's use (2,2) (2)^{2} ? 2(2) 4 = 4 ✓ 

9) Write the equation for this parabola.

This parabola opens along the negative X axis and its vertex is (3,5) which is not the origin. So, the general equation we need to use is (y  k)^{2} = 4c(x  h) (h,k) = (3, 5) and the directrix, reading the graph, is 0.25 units form the vertex. c is therefore 0.25. Substituting all of these values into the general equation we get: (y  (5))^{2} = 4(0.25)(x  (3)) (y  5)^{2} =  (x + 3) since: 4 • 0.25 = 1
check: using (7,7) ((7)  5)^{2} ? ((7) + 3) 2^{2} ?  (4) 4 = 4 ✓ 
Examples: