Conic SectionsThe Focus-Directrix Equation of a Parabola
The Standard Orientation Parabola Equation
The basic definition of a parabola is a locus of all points such that the point in the locus is equidistant to a fixed point and to a line, called the directrix. The image below shows the parts of a parabola.
The parabola in the image has its vertex at point V and its focus at point F(a, b). The focal distance, f, is the distance from the vertex to the focus. This distance is also equal to the distance from the vertex to the directrix since V is a point on the parabola.
By definition, a point, P(x, y), on the parabola is equidistant to the focus and to point D on the directrix. Therefore, FP = PD. Also, by definition FV equals VC because V is on the parabola. Therefore, ED also equals f because of parallel lines.
The equation of this parabola is easy to state. Using the distance formula, we have the length of FP as: \(d = \sqrt{(x-a)^2 + (y-b)^2}\).
The length ED equals f. Thus, we have PD = y – b + 2f.
Equating the two equal distances to each other, we have our parabola equation defined by the focal point F and the directrix: \(\sqrt{(x-a)^2 + (y-b)^2} = y - b + 2f\). Squaring both sides to remove the radical and simplifying gives us our parabola equation in focus-directrix form: \((x-a)^2 + (y-b)^2 = (y - b + 2f)^2\). The focus is at (a, b) and the directrix equation is y – b + 2f = 0 or y = b – 2f. We can also simplify further to put the equation in general form.
The equation of a parabola whose focus is at (a, b) and whose focal distance (the distance between the vertex and the focus) is f and whose directrix is given by the equation y = b – 2f is given by the equation: \(x^2 -2ax -4fy + a^2 + 4bf -4f^2 = 0\) or \(y = \frac{1}{4f}(x-a)^2+b-f\).
This equation is a little different than what we are used to seeing. However, we have kept the variables a and b that locates our focus and f that gives us our focal distance, which is also the distance of the directrix from the vertex, intact.
The equation becomes simpler if our vertex is on the x-axis. In that case, b = f and the equation becomes: \(x^2 -2ax -4by + a^2 = 0\) or \(y = \frac{1}{4b}(x-a)^2\). And if the vertex is at the origin, the equation simply becomes \(y = \frac{1}{4b}x^2\). In this cases, the variable b represents the distance between the vertex and the directrix which is also the distance between the vertex and the focus.
The General Equation of a Parabola
Now, what we wish to do is find the general equation of a parabola like we did above, but for any angle of rotation. And we want this equation based on the location of the focus and the equation of the line defining the directrix.
Once we find this equation, our goal is to define the coefficients of the general equation of conics \(A'x^2 + B'xy + C'y^2 +D'x +E'y +F' = 0\) in terms of the variables of the focus and the directrix. Therefore, given any parabola in any orientation, we want to be able to find its focus and directrix based on the coefficients.
For the remainder of the document, we will eliminate the primes in the variables for rotated conics to make it easier to follow the equations.
Based on the image above, we can state that FP = PD by definition of a parabola. The distance between point F and P is easy to state: \(d = \sqrt{(x-a)^2 + (y-b)^2}\).
The distance between P and D is the perpendicular distance between a point and a line or the shortest distance between a point and a line. I will direct readers to read this page if they are not familiar with calculating the distance between a point and a line: Shortest Distance between a Point and a Line. Knowing this formula greatly reduces the calculations here. Basically, the distance between a point (x0, y0) and a line, –mx + y – b = 0, is \(\frac{|mx_0-y_0+b|}{\sqrt{1+m^2}}\).
Applying this formula for the distance PD, we get (after changing some variables to match our line equation here): \(d = \frac{|mx-y+c|}{\sqrt{1+m^2}}\). We are keeping x and y general since they are the coordinates on the parabola. Equating these two equal distances and squaring both sides gives us our parabola equation in terms of the focus and the directrix: \((x-a)^2 + (y-b)^2 = \frac{(mx-y+c)^2}{1+m^2}\).
The equation of a parabola with its focus at (a, b) and its directrix line, y = mx + c is given by \((x-a)^2 + (y-b)^2 = \dfrac{(mx-y+c)^2}{1+m^2}\).
And we are done here. We have an equation that allows us to define our parabola by choosing our own focus point and directrix equation. The parabola will be in any orientation we want based on the slope of our directrix line. Below is a Geogebra activity that will allow you to play with the focus point and the line to see how the parabola rotates around and shifts positions. You can also move the point P on the parabola to see that the ratio remains equal to 1.
Equating With the General Equation of Conics
Well, we are not done yet. Our goal was to put the above equation into the form of the general equation of conics so we can find the focus and the directrix based on the coefficients of the general equation. We need to multiply out the equation above.
(i) \(x^2 - 2ax + a^2 + y^2 - 2by + b^2 = \) \( \frac{1}{1+m^2}(m^2x^2 - 2mxy + 2mcx + y^2 - 2cy + c^2)\)
(ii) \(x^2 - 2ax + a^2 + y^2 - 2by + b^2 = \) \(\frac{m^2}{1+m^2}x^2 - \frac{2m}{1+m^2}xy + \frac{2m}{1+m^2}cx + \frac{1}{1+m^2}y^2 - \frac{2c}{1+m^2}y + \frac{1}{1+m^2}c^2\)
(iii) \(\left(1- \frac{m^2}{1+m^2}\right)x^2 + \frac{2m}{1+m^2}xy + \) \( \left( 1 - \frac{1}{1+m^2}\right)y^2 - \left(2a + \frac{2mc}{1+m^2}\right)x - \) \( \left(2b - \frac{2c}{1+m^2}\right)y + a^2 + b^2 - \) \( \frac{c^2}{1+m^2} = 0\)
(iv) \(\left(\frac{1}{1+m^2}\right)x^2 + \frac{2m}{1+m^2}xy + \) \( \left(\frac{m^2}{1+m^2}\right)y^2 - \left( \frac{(1+m^2)(2a) + 2mc}{1+m^2}\right)x - \) \( \left(2b - \frac{2c}{1+m^2}\right)y + a^2 + b^2 - \) \( \frac{c^2}{1+m^2} = 0\)
(v) \(x^2 + (2m)xy + (m^2)y^2 - [(1+m^2)(2a) +2mc)]x -\) \( 2[(1+m^2)b - c]y + (1+m^2)(a^2 + b^2) - c^2 = 0\) (Multiplying both sides by (1+m²))
Equation (v) is in the general \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) form. We can equate the coefficients with the values in (v) and solve for a, b, c, and m in terms of A, B, C, D, E, and F so that we can determine the focus and directrix of any parabola given in general form.
Since the coefficient of the x² term is 1, we need to scale our general equation so that the x² term is 1: \(x^2 + \frac{B}{A}xy + \frac{C}{A}y^2 + \frac{D}{A}x + \frac{E}{A}y + \frac{F}{A} = 0\). We will use this equation going forward.
Equation (v) gives us the following equalities with the coefficients:
(1) \(2m = \frac{B}{A}\)
(2) \(m^2 = \frac{C}{A}\)
(3) \(-2[(1 + m^2)a + mc] = \frac{D}{A}\)
(4) \(-2[(1 + m^2)b - c] = \frac{E}{A}\)
(5) \((1 + m^2)(a^2+b^2) - c^2 = \frac{F}{A}\)
The Slope of the Directrix
From equations (1) and (2), we have two variables that give us the slope in terms of the coefficients: the xy term and the y² term. Therefore, \(2m = \frac{B}{A}\) and \(m^2 = \frac{C}{A}\). So, our two formulas for the slope are: \(m = \frac{B}{2A}\) and \(m = \pm \sqrt{\frac{C}{A}}\). The sign of the radical is the sign of the B' term. If B is negative, then \(m = -\sqrt{\frac{C}{A}}\). If B is positive, then \(m = \sqrt{\frac{C}{A}}\).
The above equations also mean that \(\frac{B}{2A} = \pm \sqrt{\frac{C}{A}}\). This is simply the determinant of the parabola, \(B^2 - 4AC = 0\), in a different form. It is best to use \(\frac{B}{2A}\) for the slope because the sign goes with the B coefficient.
Therefore, we now have the slope of the directrix in terms of the coefficients.
The slope, m, of the directrix of the parabola \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) is given by \(m = \dfrac{B}{2A}\).
The Y-Intercept of the Directrix
Equations (3), (4), and (5) are a system of equations of a, b, and c. It is easiest to solve for c by isolating a and b in equations (3) and (4), then substituting in (5) and solving for c. Then, we can solve for a and b.
From (3), we have \(a = -\frac{\frac{D}{2A} + mc}{1+m^2}\).
From (4), we have \(b = \frac{-\frac{E}{2A} + c}{1+m^2}\).
The algebra of substituting these values in (5) is messy so we will skip the steps here since it is just algebra. The intermediate equation we get for c is: \(c = \frac{4(m^2+1)F - (D^2+E^2)}{4mD - 4E}\), where the coefficients have not been scaled by 1/A.
After scaling and substituting –B/(2A) for m, the y-intercept, c, of the directrix, is given by the equation: \(c = \frac{B^2F + 4A^2F - A(D^2 + E^2)}{2ABD - 4A^2E}\). If we replace \(B^2\) with \(4AC\), then we get \( \frac{4CF + 4AF - D^2 - E^2}{2BD - 4AE}\).
What a mess to find the y-intercept. You may be wondering if this is even worthwhile. Nevertheless, it is possible to find the equation of the directrix based on the coefficients of the general parabola equation. Since we’ve come this far, let’s march forward and get it done.
The y-intercept, c, of the directrix of the parabola \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) is given by \(c = \frac{B^2F + 4A^2F - A(D^2 + E^2)}{2ABD - 4A^2E} =\) \( \frac{4CF + 4AF - D^2 - E^2}{2BD - 4AE}\).
The X-Coordinate of the Focus
The variable a is the x-coordinate of the focus. When we solve for a, we get the following intermediate equation: \(a = \frac{-mD^2 + 2DE + mE^2 - 4m(m^2+1)F}{4(m^2+1)(mD-E)}\), which has not been scaled by 1/A. After doing that and substituting –B/(2A) for m, we get the following: \(a = \frac{-ABD^2 + 4A^2DE + ABE^2 - B^3F - 4A^2BF}{4A^2BD - 8A^3E + B^3D - 2AB^2E}\). If we replace \(B^2\) with \(4AC\), we get \(a = \frac{-ABD^2 + 4A^2DE + ABE^2 - 4ACBF - 4A^2BF}{4A^2BD - 8A^3E + 4ACBD - 8A^2CE} = \) \( \frac{-BD^2 + 4ADE + BE^2 - 4BCF - 4ABF}{4ABD - 8A^2E + 4BCD - 8ACE} \).
The x-coordinate of the focus, a, is given by the equation: \(a = \frac{-ABD^2 + 4A^2DE + ABE^2 - B^3F - 4A^2BF}{4A^2BD - 8A^3E + B^3D - 2AB^2E} = \) \( \frac{-BD^2 + 4ADE + BE^2 - 4BCF - 4ABF}{4ABD - 8A^2E + 4BCD - 8ACE} \).
Well, this is even more complex than the y-intercept. The expression for a however verifies.
The Y-Coordinate of the Focus
The variable b is the y-coordinate of the focus. When we solve for b, we get the following intermediate equation: \(b = \frac{-D^2 + E^2 - 2mDE + 4(m^2+1)F}{4(m^2+1)(mD-E)}\), which has not been scaled by 1/A. After doing that and substituting –B/(2A) for m, we get the following: \(b = \frac{-2A^2D^2 + 2A^2E^2- 2ABDE + 8A^3F + 2AB^2F}{4A^2BD - 8A^3E + B^3D - 2AB^2E}\). If we replace \(B^2\) with \(4AC\), we get \(b = \frac{-2A^2D^2 + 2A^2E^2- 2ABDE + 8A^3F + 8A^2CF}{4A^2BD - 8A^3E + 4ACBD - 8A^2CE} = \) \( \frac{-AD^2 + AE^2 - BDE + 4A^2F + 4ACF}{2ABD - 4A^2E + 2BCD - 4ACE} \).
The y-coordinate of the focus, b, is given by the equation: \(b = \frac{-2A^2D^2 + 2A^2E^2- 2ABDE + 8A^3F + 2AB^2F}{4A^2BD - 8A^3E + B^3D - 2AB^2E} = \) \( \frac{-AD^2 + AE^2 - BDE + 4A^2F + 4ACF}{2ABD - 4A^2E + 2BCD - 4ACE} \).
Again, we get a behemoth formula for the y-coordinate and it checks. The numbers come out huge when evaluating with real coefficients.
You may notice the denominators are the same for a and b.
Well, I’m not going to put all of a, b, c, and m into one equation of our parabola because the formulas are so long. However, we have completely solved for these variables in terms of the coefficients of the parabola. Therefore, we can find the focus and the directrix equation for any given parabola written in general equation form.
The Angle of Rotation
Did you notice something while we were solving for the slope... particularly in equation (1)?
The slope of the directrix is the angle the line creates with the x-axis. It may seem we can calculate the angle of rotation, θ, of the parabola using the formula \(\theta = \tan^{-1}(m) = \tan^{-1}\frac{B}{2A}\). However, this does not always work when some coefficients have negative values.
We need to investigate why.
The Vertex
Finding the vertex in terms of the coefficients is very complicated based only on the variables because it is a multistep process. There are two ways of finding the vertex.
(1) First, we can find the distance between the focus and the directrix. Then half of this distance would be the distance between the focus and the vertex. Then, we would find the sine and cosine of the angle of rotation to find the x and y displacements. Lastly, we would subtract or add the displacements from the focus coordinates depending on where the vertex is in relation to the focus.
(2) The second and easier method is to find the equation of the line of symmetry (its slope is the negative inverse of the directrix and the focus is a point on the line). Then, find the intersection of the parabola and this line. It would have only one solution.
The second method is by far the easier method. We will employ both methods in 2 different examples here.
Examples
Let’s do a problem and see if we are getting the correct answers. We will check with Goegebra.
Example 1
Problem 1: Find the slope of the directrix, the angle of rotation of the parabola, and the focus of the parabola \(9x^2+12xy+4y^2 - 72x+108y-36=0\). Then, find the focus.
Solution: Let’s find the slope first, which is quite simple. The slope is B/(2A). So, the slope is \(m = \frac{12}{2\cdot 9} = \frac{2}{3}\).
Angle of rotation: The angle of rotation formula is: \(\theta = \frac{1}{2}\cot^{-1}\frac{A-C}{B}\). So, our angle of rotation is \(\theta = \frac{1}{2}\cot^{-1}\frac{5}{12} = 33.69^{\circ}\).
The y-intercept: Now, we find the y-intercept. The formula is \(c = \frac{B^2F + 4A^2F - A(D^2 + E^2)}{2ABD - 4A^2E}\). Plugging in the values, we get: \(c = \frac{(12)^2(-36) + 4(9)^2(-36) - (9)((-72)^2 + (108)^2)}{2(9)(12)(-72) - 4(9)^2(108)} = \frac{10}{3}\). That has to be some complicated math with only numbers.
The x-coordinate of the focus: Now, let’s plug in numbers into the formula \(a = \frac{-ABD^2 + 4A^2DE + ABE^2 - B^3F - 4A^2BF}{4A^2BD - 8A^3E + B^3D - 2AB^2E}\) to get our x-coordinate: \(a = \frac{-(9)(12)(-72)^2 + 4(9)^2(-72)(108) + (9)(12)(108)^2 - (12)^3(-36) - 4(9)^2(12)(-36)}{4(9)^2(12)(-72) - 8(9)^3(108) + (12)^3(-72) - 2(9)(12)^2(108)} = \frac{16}{13}\).
The above math was a doozy. I got a value of 1.230769... as a continued fraction with 230769 digits repeating. Turned out to be 16/13 after manually reducing it.
The y-coordinate of the focus: The formula is \(b = \frac{-2A^2D^2 + 2A^2E^2- 2ABDE + 8A^3F + 2AB^2F}{4A^2BD - 8A^3E + B^3D - 2AB^2E}\). Pluggin in: \(b = \frac{-2(9)^2(-72)^2 + 2(9)^2(108)^2- 2(9)(12)(-72)(108) + 8(9)^3(-36) + 2(9)(12)^2(-36)}{4(9)^2(12)(-72) - 8(9)^3(108) + (12)^3(-72) - 2(9)(12)^2(108)} = -\frac{24}{13}\).
We now have all of our variables solved to write our equation of the parabola: \(\left(x-\frac{16}{13}\right)^2 + \left(y + \frac{24}{13}\right)^2 = \frac{9}{13}\left({\frac{2}{3}x - y + \frac{10}{3}}\right)^2\). The focus is at \(\left( \frac{16}{13}, -\frac{24}{13} \right)\) and the equation of the directrix is \(\frac{2}{3}x - y + \frac{10}{3} = 0\) or \(y = \frac{2}{3}x + \frac{10}{3}\).
The vertex: To find the vertex, we need to find the distance between the focus and the directrix, and then we take half the distance to find the vertex point. We can use the point-line distance formula to find the distance. The equation of the directrix is \(\frac{2}{3}x - y + \frac{10}{3} = 0\) or \(2x - 3y + 10 = 0\). The distance between the focus and the directrix, d, is: \(d = \frac{\left|2(16/13) - 3(-24/13) + 10\right|}{\sqrt{2^2+4^2}} =\) \( \frac{234/13}{\sqrt{13}} = \frac{18}{\sqrt{13}} \approx 4.99....\). Half of this distance is \(\frac{9}{\sqrt{13}}\).
We need to find the horizontal and vertical displacements to find the coordinates of the vertex. The segment connecting the vertex and the focus has a slope of –3/2. Therefore, \(\tan\theta = -\frac{3}{2}\). From this we can find the sine and cosine: \(\sin\theta = \frac{3}{\sqrt{13}}\) and \(\cos\theta = \frac{2}{\sqrt{13}}\). The sign does not matter because we need to subtract the horizontal distance and add the vertical distance based on the location of the vertex relative to the focus.
The horizontal displacement is \(-\frac{2}{\sqrt{13}}\cdot \frac{9}{\sqrt{13}} = -\frac{18}{13}\) and the vertical displacement is \(+\frac{3}{\sqrt{13}}\cdot \frac{9}{\sqrt{13}} = +\frac{27}{13}\). Therefore, the location of the vertex is \(\left( \frac{16}{13} - \frac{18}{13}, -\frac{24}{13} + \frac{27}{13} \right) = \left( -\frac{2}{13}, \frac{3}{13} \right)\). The figure below shows our parabola with its focus, vertex, and the directrix. A point B on the parabola is shown so we can see the parabola ratio is equal to 1.
The vertex again: Here’s an easier method to find the vertex. The vertex is the intersection of the line of symmetry and the parabola. The line of symmetry has a slope that is the negative inverse of the directrix and the focus is on this line. The equation of the line of symmetry for our example above is: \(y + \frac{24}{13} = -\frac{3}{2}(x-(16/13))\). This is the same as \(y = -\frac{3}{2}x\). Looks like our line of symmetry passes through the origin.
We can substitute this into the parabola equation and solve for x:
(i) \(9x^2+12x\cdot \left(-\frac{3}{2}x\right) + 4\cdot \left(-\frac{3}{2}x\right)^2 - 72x + 108\cdot \left(-\frac{3}{2}x\right)-36=0\)
(ii) \(9x^2 -18x^2 + 9x^2 - 72x - 162x -36=0\)
(iii) \(-234x = 36\) (That was a convenient simplication. Does that happen every time?)
(iv) \(x = -\frac{36}{234} = -\frac{2}{13}\)
The y value is trivial: \(y = -\frac{3}{2}\cdot -\frac{2}{13} = \frac{3}{13}\).
The standard form of the parabola in this example is \(13x^2+\frac{468}{\sqrt{13}}y-36=0\). The parabola above was also used as an example in Rotation of Parabolas page. We found its vertex in a different way.
Example 2
Find the slope of the directrix, the angle of rotation, the focus point, the vertex of the parabola \(9x^2-24xy+16y^2-130x+90y=0\).
The slope: The slope of the directrix is: \(m = \frac{B}{2A} = \frac{-24}{2\cdot 9} = -\frac{4}{3}\). Because the slope is negative, we know that the parabola opens up in Quadrant I or III. This one opens up in Quadrant I.
The angle of rotation: The angle of rotation is given by \(\frac{1}{2}\cot^{-1}\frac{-7}{-24} = 36.87^{\circ}\).
The Y-Intercept: Let’s find the y-intercept to complete the directrix equation: \(c = \frac{(-24)^2(0) + 4(9)^2(0) - (9)((-130)^2 + (-90)^2)}{2(9)(-24)(-130) - 4(9)^2(90)} = -\frac{25}{3}\).
The directrix equation: Now that we have the slope and y-intercpet, we have our directrix equation: \(y = -\frac{4}{3}x - \frac{25}{3}\).
The coordinates of the focus: Now, we will use a shortcut to find the focus. Instead of using our long complicated formula, let’s just use equations (3) and (4). We’ll start with a using equation (3):
(i) \(-2[(1 + m^2)a + mc] = \frac{D}{A}\)
(ii) \(-2[(1 + (-4/3)^2)a + (-4/3)(-25/3)] = \frac{-130}{9}\)
(iii) \(-2\left(\frac{25}{9}a+\frac{100}{9}\right)=\frac{-130}{9}\)
(iv) \(\frac{25}{9}a=-\frac{35}{9}\)
(v) \(a=-\frac{7}{5}\)
We found our x-coordinate in an easier way. We’ll find the y-coordinate using equation (4).
(i) \(-2[(1 + m^2)b - c] = \frac{E}{A}\)
(ii) \(-2[(1 + (-4/3)^2)b - (-25/3)] = \frac{90}{9}\)
(iii) \(-2\left(\frac{25}{9}b+\frac{25}{3}\right)=10\)
(iv) \(\frac{25}{9}b=-\frac{40}{3}\)
(v) \(b=-\frac{24}{5}\)
Our focus is located at \(\left( -\frac{7}{5}, -\frac{24}{5} \right)\). And boy! Was that a lot easier than using the coefficient formulas!
Our equation of the parabola is: \(\left(x+\frac{7}{5}\right)^2+\left(y+\frac{24}{5}\right)^2 = \frac{9}{25}\left(\frac{4}{3}x+y+\frac{25}{3}\right)^2\).
The vertex: We will find the vertex by finding the intersection of the line of symmetry with the parabola. The line of symmetry has a slope of 3/4. The equation is \(y + \frac{24}{5} = \frac{3}{4}\left(x+\frac{7}{5}\right)\) or \(y = \frac{3}{4}x - \frac{15}{4}\). Plugging in the y into the parabola equation:
(i) \(9x^2-24x\left(\frac{3}{4}x-\frac{15}{4}\right)+16\left(\frac{3}{4}x-\frac{15}{4}\right)^2-130x+90\left(\frac{3}{4}x-\frac{15}{4}\right) = 0\)
(ii) \(9x^2-18x^2+90x+ 9x^2-90x + 225 -130x+\frac{135x}{2}-\frac{675}{2} = 0\)
(iii) \(-\frac{125}{2}x - \frac{225}{2} = 0\)
(iv) \(x = -\frac{9}{5}\)
To find y, we substitute x into the line of symmetry equation: \(y = \frac{3}{4}\cdot -\frac{9}{5} - \frac{15}{4} = -\frac{51}{10}\). Therefore, our vertex is at \(\left(-\frac{9}{5}, -\frac{51}{10}\right)\). Again, equation (ii) became friendly with the x² term cancelling out.
The figure below shows a zoom in of our parabola at the vertex.
The above figure is zoomed in to the vertex and the focus. It may appear that the vertex and the focus are far apart. But when zooming out, you can see how close the focus is to the vertex.
The equation of this parabola when rotated to standard form is \(25y^2-50x+150y = 0\) or \(x = \frac{1}{2}y^2+3y\).
Last Word
Did you notice something else about these equations? Anything at all? The slope, the y-intercept, the vertex coordinates, and the focus coordinates are rational numbers. If our parabola has rational coefficients, this will always be true. If the parabola has irrational coordinates, this will not be true.
The distance between the focus and the vertex was irrational. When calculating the displacement, because the trigonometric values were irrational, they canceled out in the horizontal and vertical displacements. So we ended up with rational horizontal and vertical displacements.