• Algebra > Parabola Rotation

Rotation of Parabolas

Rotation of General Parabola to Standard Position

The general form of a conic is \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). This conic could be a circle, parabola, ellipse, or a hyperbola in any orientation, meaning it could be rotated so that the directrix is not vertical or horizontal but at an angle.

We can identify the conic based on A, B, and C, however. As discussed on the Conic Sections page, the following determinants allows us to determine the conic:

(1) If \(B^2 - 4AC = 0\), the conic is a parabola.

(2) If \(B^2 - 4AC < 0\), the conic is an ellipse.

(3) If \(B^2 - 4AC > 0\), the conic is a hyperbola.

From (1) above, we can see that the determinant of the parabola is an equality, not an inequality. That means \(B^2 = 4AC\) if the conic is a parabola. (This also means that A and C must have the same sign and this product is always positive.) We can substitute B² for 4AC or vice versa. This makes the parabola unique among the conics, because this equality allows us to simplify the coefficients of the general parabola rotated back to its standard orientation, provided we know it is a parabola by checking the determinant.

Before we simplify the formulas for the coefficients, let’s revisit the rotation formula and the angle of rotation and the trigonometry that follows.

The Rotation Formula for General Conics

If a conic defined by a general equation is rotated by an angle θ, then the following formula for the rotated conic is the result:

In this equation, we need to find the critical angle θ that will put our conic in standard orientation. That angle is the angle that makes the xy term 0. That angle can be found by simple algebra and is given next.

The Trigonometry of the Angle of Rotation

The explanation of rotation angle that rotates a conic in standard orientation has been covered in the Conic Sections page. To understand which sign is used that follows, please review what is explained in the Conic Sections page.

The angle of rotation, θ, is \(\theta = \frac{1}{2}\cot^{-1}\left(\frac{A-C}{B}\right)\). If \(\frac{A-C}{B} \) is negative, the angle of rotation is negative and vice versa. If A = C, the angle of rotation is 45°.

We can create a right triangle with the legs whose values are A – C and B and find the hypotenuse. From that we can calculate the sine and cosine of the rotation angle. We will use the absolute values of the lengths because only the absolute length is used to calculate the hypotenuse.

In Figure 1, the hypotenuse is \( \sqrt{|A-C|^2 + |B|^2} \). But since \(B^2 = 4AC \) for a parabola, the hypotenuse is \( \sqrt{A^2 - 2AC + C^2 + 4AC} = |A+C| \) for a parabola.

From Figure 1, we can determine sine and cosine of θ: \( \cos(2\theta) = \frac{|A-C|}{|A+C|} \). Using the half-angle identity: \( \cos(\theta) = \sqrt{\frac{1}{2} + \frac{1}{2}\cdot\frac{|A-C|}{|A+C|}} = \sqrt{\frac{|A+C| + |A-C|}{2|A+C|}} \).

Remember that the sign of sine depends on whether the angle of rotation is positive or negative.

Since both A and C must have the same sign, the quantity \( |A+C| + |A-C| \) will always equal twice the greater value of |A| or |C|. The formulas above can be simplified if we know which is greater. If |A| > |C|, then \( \cos\theta = \sqrt{\dfrac{|A|}{|A+C|}}\), and if |C| > |A|, then \( \cos\theta = \sqrt{\dfrac{|C|}{|A+C|}}\). The quantity \( |A+C| - |A-C| \) will always equal twice the smaller value of |A| or |C|. Therefore, if |A| < |C|, then \( \sin\theta = \pm \sqrt{\dfrac{|A|}{|A+C|}}\), and if |C| < |A|, then \( \sin\theta = \pm \sqrt{\dfrac{|C|}{|A+C|}}\).

In summary, the numerators in sine and cosine are switched. And remember, top sign for positive angle of rotation and bottom sign for negative angle of rotation.

Another point: if |A| > |C|, then the parabola will rotate so that its line of symmetry is parallel with the y-axis. Conversely, if |C| > |A|, then the parabola will rotate so that its line of symmetry is parallel with the x-axis. The angle of rotation is always the minimum angle required to render the conic in standard position, meaning the angle of rotation is greater than 0° or less than 45°. If A = C, then the angle of rotation is 45° and the conic rotates so that its line of symmetry is parallel to the x-axis - meaning the axes rotated in the positive direction or counterclockwise.

Simplifying the Rotation Formula for a Parabola

Using the rotation formula for conics stated above and the trigonometric values we derived above, we can simplify the coefficient formulas for the parabola.

So, let’s derive the simpler formulas of the coefficients one-by-one.

To evaluate A', I made a table showing the signs of A and C and the rotation angle to show how each scenario gives us the same value. The variables A and C can be inherently positive or negative, without showing the sign. Therefore, if A is negative, we will show these values as –|A| and –|C|. We will show the positive values as |A| and |C|. Remember that AC is always positive because they must have the same sign.

The formula for A' is \(A' = A\cos^2\theta + \dfrac{1}{2}B\sin(2\theta)+C\sin^2\theta \). If |A| > |C|, the evaluations for the different scenarios are in the table below.

These four are the only scenarios and they all add to A + C.

In rows 3 and 5, –|A + C| became A + C because they were negative already, so the sign is not necessary.

If |A| > |C|, then C' must equal 0 since only either A' or C' can be non-zero. The formula for C' is \(C' = A\sin^2\theta -\dfrac{1}{2}B\sin(2\theta) + C\cos^2\theta \). We have the following 4 scenarios and evaluations for C':

Therefore, C' always evaluates to 0. And it must be the case since A' evaluated to a non-zero number. Both cannot be non-zero for a standard orientation parabola. Therefore, we have the following evaluations for A' and C'.

(i) \( A' = \left(A\cos^2\theta + \dfrac{1}{2}B\sin(2\theta)+C\sin^2\theta\right) = A+C\)

(ii) \(C' = \left(A\sin^2\theta -\dfrac{1}{2}B\sin(2\theta) + C\cos^2\theta\right ) = 0\)

In similar fashion, we can also show that if |C| > |A|, the opposite happens. Therefore,

(i) \( A' = \left(A\cos^2\theta + \dfrac{1}{2}B\sin(2\theta)+C\sin^2\theta\right) = 0\)

(ii) \(C' = \left(A\sin^2\theta -\dfrac{1}{2}B\sin(2\theta) + C\cos^2\theta\right ) = A+C\)

Now for D and E. If |A| > |C|, then:

(iii) \(D' = (D\cos\theta + E\sin\theta) = \) \(\dfrac{D\sqrt{|A|} \pm E\sqrt{|C|}}{\sqrt{|A+C|}}\)

and if |C| > |A|, then:

(iii) \(D' = (D\cos\theta + E\sin\theta) = \) \(\dfrac{D\sqrt{|C|} \pm E\sqrt{|A|}}{\sqrt{|A+C|}}\)

If |A| > |C|, then:

(iv) \(E' = (E\cos\theta - D\sin\theta) = \) \(\dfrac{E\sqrt{|A|} \mp D\sqrt{|C|}}{\sqrt{|A+C|}}\)

and if |C| > |A|, then:

(iv) \(E' = (E\cos\theta - D\sin\theta) = \) \(\dfrac{E\sqrt{|C|} \mp D\sqrt{|A|}}{\sqrt{|A+C|}}\)

(v) \(F' = F\)

If A < C, alternatively, we use a trick: we can switch the x and y to get the inverse of the conic, apply the formulas for |A| > |C|, then switch the x and y again. That will give us our desired conic in standard orientation.

Let’s put this together in one equation.

The Vertex

The first equation above can be written as a function form: \(y = -\frac{\sqrt{(A+C)^3}}{E\sqrt{A} \mp D\sqrt{C}}x^2 - \frac{D\sqrt{A} \pm E\sqrt{C}}{E\sqrt{A} \mp D\sqrt{C}}x - \frac{F\sqrt{A+C}}{E\sqrt{A} \mp D\sqrt{C}}\).

The form above allows to find the vertex of the standard parabola based on the coefficients of the original rotated parabola. For a parabola \(y=ax^2+bx+c\), the vertex is located at \(\left(-\frac{b}{2a}, \frac{-b^2+4ac}{4a}\right)\). Doing the cumbersome math, we get the following for the focus of the standard orientation parabola based on the coefficients of the rotated parabola.

If the conic \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) is determined to be a parabola and |A| > |C|, then its vertex in standard position is located at: \(\left(-\frac{D\sqrt{|A|} \pm E\sqrt{|C|}}{2\sqrt{|A+C|^3}} , \frac{(D\sqrt{|A|} \pm E\sqrt{|C|})^2 - 4F(A+C)^2}{4\sqrt{|A+C|^3}(E\sqrt{|A|} \mp D\sqrt{|C|})} \right)\). (Remember to apply the correct sign based on the angle of rotation.)

Rotation Examples

Example 1

Problem: Rotate the following parabola to its standard form and find the angle of rotation: \(16x^2 + 24xy + 9y^2 -130x +90y = 0\). Also, find the vertex of this rotated parabola.

Solution: Let’s make sure it is a parabola by checking its determinant: \(24^2 - 4(16)(9) = 576-576 = 0\). The determinant is 0, so it is a parabola.

First, we find the rotation angle because the sign depends on the rotation angle.

The angle of rotation is \(\theta = \frac{1}{2}\cot^{-1}\left(\frac{16-9}{24}\right) = \frac{1}{2}\cot^{-1}\frac{7}{24} \approx 36.89^{\circ}\). The rotation angle is positive and this is a counterclockwise rotation angle from standard orientation. Therefore, we will use the positive sign for our x coefficient and a negative sign for our y coefficient.

Lastly, |A| > |C|, so we can use the formula which lacks the y² term because the parabola will rotate in the direction that places its line of symmetry parallel to the y-axis.

Let’s put this in our rotation formula: \((A+C)x^2 + \left(\frac{D\sqrt{A}+E\sqrt{C}}{\sqrt{A+C}}\right)x \text{ } + \) \( \left(\frac{E\sqrt{A}-D\sqrt{C}}{\sqrt{A+C}}\right)y + F = 0 \)

(i) \((16+9)x^2 + \left(\frac{-130\sqrt{16} + 90\sqrt{9}}{\sqrt{16+9}}\right)x \text{ } + \) \( \left(\frac{90\sqrt{16} -(-130)\sqrt{9}}{\sqrt{16+9}}\right)y + 0 = 0 \)

(ii) \(25x^2 - 50x + 150y = 0 \) or \(x^2 - 2x + 6y = 0\) or \(y=-\frac{1}{6}x^2 + \frac{1}{3}x\)

Vertex: To find the vertex of the rotated parabola, we will rotate the vertex of the standard parabola by the angle we found above in the counterclockwise direction.

The x coordinate of the vertex of the standard parabola is \(\frac{2}{2\cdot 1} = 1\). Using this value, our y coordinate is \(-\frac{(1)^2}{6} + \frac{1}{3}(1) = \frac{1}{6}\). So, our standard focus is at (1, 1/6).

To rotate points counterclockwise, we use the formulas: \(x' = x\cos\theta - y\sin\theta\) and \( y' = x\sin\theta + y\cos\theta\). We need to find the sine and cosine of the angle of rotation. We can construct a right triangle with sides 7 and 24 and the angle 2θ to find that the hypotenuse is 25. Therefore, \(\cos(2\theta) = \frac{7}{25}\). Using the half-angle identity: \(\cos\theta = \sqrt{\frac{1}{2} + \frac{1}{2}\cdot\frac{7}{25}} = \sqrt{\frac{32}{50}} = \frac{4}{5}\). In a similar manner, the sine is \(\sin\theta = \sqrt{\frac{1}{2} - \frac{1}{2}\cdot\frac{7}{25}} = \sqrt{\frac{18}{50}} = \frac{3}{5}\).

Now, we apply the rotation formulas: \(x: (1)\cdot\frac{4}{5}-\frac{1}{6}\cdot\frac{3}{5} = \frac{4}{5} - \frac{1}{10} = \frac{7}{10}\) and \(y: (1)\cdot\frac{3}{5}+\frac{1}{6}\cdot\frac{4}{5} = \frac{3}{5} + \frac{2}{15} = \frac{11}{15}\). Therefore, our vertex on the original parabola is located at (7/10, 11/15).

Figure 3 shows our two parabolas, one in standard position and the other in rotated position. The focus A has been translated to B and you can see the angle in between is about 36.87°.

When F = 0, the parabola passes through the origin. Parabolas that pass through the origin have a curious effect when rotated. They pivor about the origin point, which is an expected behavior. In the Geogebra activity above, readers are encourages to change the coefficients to this parabola’s coefficients and see the rotation effect.

The Focus: The focus of the standard parabola can be found using the formula \(-\frac{b}{2a}\) for the focal distance and off-setting that from the vertex. The focus is located at \(\left(1, -\frac{4}{3}\right)\). The focus of the original rotated parabola, after rotating the standard focus just like we did the vertex above, is located at \(\left(\frac{8}{5}, -\frac{7}{15}\right)\).

Proving the Vertex of the Rotated Parabola

When we find the vertex, how do we know if it is actually the vertex? It could be a point very close to the vertex, imperceptible to the eye.

For a parabola in standard orientation, a vertical line passes through one and only one point on the parabola. If it passes through more than one point, the line has a finite slope. The tangent at the vertex is horizontal, meaning it has a slope of 0.

Therefore, for a rotated parabola, this means a perpendicular to the tangent at the vertex would have only one intersection with the parabola. In other words, the equation of the line and the parabola have only one solution. So, for our example above, let’s find the slope at the vertex, then the slope of the line and solve the system of equations.

We will use implicit differentiation to find the slope of the tangent at (7/10, 11/15).

(i) \(16x^2+24xy+9y^2-130x+90y=0\)

(ii) \(32x + 24x + 24x\frac{dy}{dx} + 18y\frac{dy}{dx} - 130 + 90\frac{dy}{dx}=0\) (Implicit differentation)

(iii) \(\frac{dy}{dx} = \frac{-16x-12y+65}{12x+9y+45}\) (solving for the derivative and simplifying)

(iii) \(m = \frac{-16(7/10)-12(11/15)+65}{12(7/10)+9(11/15)+45} = \frac{3}{4}\) (evaluating for the slope at point (7/10, 11/15))

Now, we take the negative inverse of the slope and find the equation of the line:

(iv) \(y - \frac{11}{15} = -\frac{4}{3}(x-\frac{7}{10})\)

(v) \(y = -\frac{4}{3}x + \frac{5}{3}\)

We can substitute y in equation (v) into equation (i) and solve for the solutions. We should get only one solution, which is the vertex or the point of intersection of the line and the parabola.

(vi) \(16x^2+24x(-\frac{4}{3}x + \frac{5}{3}) + 9(-\frac{4}{3}x + \frac{5}{3})^2-130x +90(-\frac{4}{3}x + \frac{5}{3}) = 0\)

(vii) \(16x^2 - 32x^2 + 40x + 16x^2 - 40x +25 - 130x - 120x +150 = 0\)

(vii) \( -250x + 175 = 0\)

(vii) \( x = \frac{7}{10}\)

The x² term canceled out leaving only 1 solution, which is the x coordinate of the vertex.

Example 2

Problem: Rotate the parabola \(\frac{1}{4}x^2+\frac{1}{3}xy+\frac{1}{9}y^2-2x+3y-1=0\) and find its angle of rotation.

Solution: We can actually multiply the whole equation by 36 to eliminate the fractions and make the equation simpler.

(i) \(9x^2+12xy+4y^2-72x+108y-36=0\)

We need to find the angle of rotation first to determine if the rotation is positive or negative. To find the angle of rotation, we will find the cotangent of twice the angle first: \(\cot(2\theta) = \frac{9-4}{12} = \frac{5}{12}\). Therefore, 2θ = arccot(5/12) ≈ 67.38°. The angle of rotation is half that or 33.69°. Since our rotation angle is positive, we will use the positive sign in x term and negative sign in y term. Also, |A| > |C|, so the parabola will open up and down and will lack the y² term.

Note that the legs of the right triangle are 5 and 12. The third side, the hypotenuse, is 13.

Now use the rotation equation:

(ii) \((9+4)x^2 + \left(\frac{-72\sqrt{9}+108\sqrt{4}}{\sqrt{9+4}}\right)x + \) \( \left(\frac{108\sqrt{9}-(-72)\sqrt{4}}{\sqrt{9+4}}\right)y - 36 = 0 \)

(iii) \(13x^2 + \left(0\right)x + \left(\frac{468}{\sqrt{13}}\right)y - 36 = 0 \)

So, our parabola in standard orientation has the equation: \(13x^2 + \frac{468}{\sqrt{13}}y - 36 = 0 \) or \(y = \frac{\sqrt{13}}{13}-\frac{\sqrt{13}}{36}x^2\).

The vertex of the standard parabola is at \(\left(0,\frac{\sqrt{13}}{13}\right)\). To find the vertex of the original parabola, we need to find the sine and cosine. The cosine of twice the angle is 5/13, so the half-angle identity gives us: \(\cos\theta = \sqrt{\frac{1+5/13}{2}} = \frac{3}{\sqrt{13}}\) and \(\sin\theta = \sqrt{\frac{1-5/13}{2}} = \frac{2}{\sqrt{13}}\).

Now, we can rotate the vertex using our counterclockwise rotation formula: \(x: (0)\cdot\frac{3}{\sqrt{13}}-\frac{\sqrt{13}}{13}\cdot\frac{2}{\sqrt{13}} = -\frac{2}{13}\) and \(y: (0)\cdot\frac{2}{\sqrt{13}}+\frac{\sqrt{13}}{13}\cdot\frac{3}{\sqrt{13}} = \frac{3}{13} \). Therefore, our original vertex is at \(\left(-\frac{2}{13}, \frac{3}{13}\right)\).

You can see the zoomed image of vertices of both parabolas and the angle they create. The vertices are very close to each other.

Now, let’s change the sign of the xy term in Example 1 above and rotate the parabola. This will illustrate how changing one sign gives a totally different parabola.

Example 3

Problem: Rotate the parabola \(16x^2-24xy+9y^2-130x+90y=0\) to standard orientation.

Solution: We will find the angle of rotation first to determine the sign.

The angle of rotation is \(\theta = \frac{1}{2}\cot^{-1}\left(\frac{16-9}{-24}\right) = -\frac{1}{2}\cot^{-1}\frac{7}{24} \approx -36.89^{\circ}\). The rotation angle is negative and this is a clockwise rotation angle from standard orientation. Therefore, we will use the negative sign for our x coefficient and a positive sign for our y coefficient.

Let’s put this in our rotation formula: \((A+C)x^2 + \left(\frac{D\sqrt{A}-E\sqrt{C}}{\sqrt{A+C}}\right)x \text{ } + \) \( \left(\frac{E\sqrt{A}+D\sqrt{C}}{\sqrt{A+C}}\right)y + F = 0 \)

(i) \((16+9)x^2 + \left(\frac{-130\sqrt{16} - 90\sqrt{9}}{\sqrt{16+9}}\right)x \text{ } + \) \( \left(\frac{90\sqrt{16} +(-130)\sqrt{9}}{\sqrt{16+9}}\right)y + 0 = 0 \)

(ii) \(25x^2 - 158x - 6y = 0 \) or \(y=\frac{25}{6}x^2 - \frac{79}{3}x\)

The image below shows our two parabolas that look nothing like the parabola in Example 1.

Vertex: The vertex of the standard parabols is at \(\left(\frac{79}{25}, -\frac{79^2}{150}\right)\).

To find the vertex of the rotated parabola, we use \(\cos\theta = \frac{4}{5}\) and \(\sin\theta = \frac{3}{5}\). (Remember that sine is negative!)

The rotation formulas are: \(x' = x\cos\theta - y\sin\theta\) and \(y' = x\sin\theta + y\cos\theta\).

The x position is \(x: \frac{79}{25}\cdot \frac{4}{5} - \left(-\frac{79^2}{150}\right)\cdot\left(-\frac{3}{5}\right) = -\frac{16827}{750} = -22.436\).

The y position is \(\frac{79}{25}\cdot \left(-\frac{3}{5}\right) + \left(-\frac{79^2}{150}\right)\cdot\frac{4}{5} = -\frac{26386}{750} \approx 35.1813....\).

Therefore, our position of the vertex of the rotated parabola is (–22.436, 35.1813...).

Notice that the vertex is a rational number. It seems if the A and C terms are squares, then the vertex is always at rational coordinates.

Example 4

Problem: Rotate the parabola \(x^2 - 4xy +4y^2 + 3x - 4y + 1 = 0 \).

Solution: Because A is less than C, we will find the inverse, rotate, then find the inverse of the rotated parabola.

The inverse of the above parabola is \(4x^2 - 4xy + y^2 - 4x + 3y + 1 = 0\). The angle of rotation of the inverse parabola is negative because B is negative and A and C are positive. We need to apply the bottom sign.

Using the short formulas, the coefficients are:

(i) \(A' = A+C = 1 + 4 = 5\)

(ii) \(D' = \frac{D\sqrt{A} \pm E\sqrt{C}}{\sqrt{A+C}} = \) \( \frac{-4\sqrt{4} - 3\sqrt{1}}{\sqrt{4+1}} = \frac{-11}{\sqrt{5}}\)

(iii) \(E' = \frac{E\sqrt{A} \mp D\sqrt{C}}{\sqrt{A+C}} = \) \(\frac{3\sqrt{4} + -4\sqrt{1}}{\sqrt{1+5}} = \frac{2}{\sqrt{5}}\)

(vi) \(F' = F = 1\)

Our inverse standard parabola is \(5x^2 - \frac{11}{\sqrt{5}}x + \frac{2}{\sqrt{5}}y + 1 = 0 \). Therefore, our orignal rotated parabola in standard orientation is the inverse of this one or \(5y^2 - \frac{11}{\sqrt{5}}y + \frac{2}{\sqrt{5}}x + 1 = 0 \).

The angle of rotation is \(\theta = \ \frac{1}{2}\cot^{-1}\left(\frac{1-4}{-4}\right) = 26.57\). The angle of rotation of our original parabola is positive. You can see below that the axes rotated counterclockwise.

The vertex of the standard parabola is at \( \left(\frac{21}{40\sqrt{5}}, \frac{11}{10\sqrt{5}} \right) \) and it can be found by differentiation.

To find the vertex of the rotated parabola, we use the rotation formulas.

The x coordinate is \(x' = x\cos\theta - y\sin\theta = \left(\frac{21}{40\sqrt{5}}\right)\left(\frac{2}{\sqrt{5}}\right) - \left(\frac{11}{10\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}}\right) = -\frac{1}{100} \).

The y coordinate is \(y' = x\sin\theta + y\cos\theta = \left(\frac{21}{40\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}}\right) + \left(\frac{11}{10\sqrt{5}}\right)\left(\frac{2}{\sqrt{5}}\right) = \frac{109}{200} \).

The vertex of the original parabola is \( \left(-\frac{1}{100}, \frac{109}{200}\right) \) or exactly at (–0.01, 0.545).

The vertex formula: If you want to use the vertex formula stated earlier, we have to find the vertex of the inverse, then switch the x and y coordinates. The vertex formula is \(\left(-\frac{D\sqrt{A} \pm E\sqrt{C}}{2\sqrt{(A+C)^3}} , \frac{(D\sqrt{A} \pm E\sqrt{C})^2 - 4F(A+C)^2}{4\sqrt{(A+C)^3}(E\sqrt{A} \mp D\sqrt{C})} \right)\).

The inverse of our original conic is \( 4x^2 - 4xy + x^2 - 4x + 3y + 1 = 0 \) and we rotated this conic in the negative direction. The x coordinate is: \( x: -\frac{D\sqrt{A} - E\sqrt{C}}{2\sqrt{(A+C)^3}} = -\frac{-4\sqrt{4} - 3\sqrt{1}}{2\sqrt{(4+1)^3}} = -\frac{-8-3}{10\sqrt{5}} = \frac{11}{10\sqrt{5}} \).

The y-coordinate is \(\frac{(D\sqrt{A} - E\sqrt{C})^2 - 4F(A+C)^2}{4\sqrt{(A+C)^3}(E\sqrt{A} + D\sqrt{C})} = \frac{(-4\sqrt{4} - 3\sqrt{1})^2 - 4(1)(4+1)^2}{4\sqrt{(4+1)^3}(3\sqrt{4} + (-4)\sqrt{1})} = \frac{11^2-4\cdot25}{40\sqrt{5}} = \frac{21}{40\sqrt{5}}\).

For the actual parabola, then, the vertex is located at the switched x and y coordinates or \( \left(\frac{21}{40\sqrt{5}}, \frac{11}{10\sqrt{5}} \right) \), which is what we got above. This substantiates our vertex formula as correct.

The focus: Let’s find the focus now. We will find the focus of the standard parabola, then apply the rotation formulas, which is the easier method. Since the parabola opens left, the focus is to the left of the vertex. We need to subtract the focal distance. For the parabola \(x = ay^2 + bx +c\), the focal distance is \( \frac{1}{4a}\). Our parabola \(5y^2 - \frac{11}{\sqrt{5}}y + \frac{2}{\sqrt{5}}x + 1 = 0 \) in that form is \( x = -\frac{\sqrt{5}}{2}y^2+\frac{11}{2}+\frac{\sqrt{5}}{2}\). Therefore, the focal distance is \(\frac{1}{(4)\left(-\frac{5\sqrt{5}}{2}\right)} = -\frac{1}{10\sqrt{5}} \). We need to subtract the absolute distance from the x coordinate of the vertex and keep the y coordinate the same. Therefore, the focus is located at \( \left(\frac{21}{40\sqrt{5}} - \frac{1}{10\sqrt{5}}, \frac{11}{10\sqrt{5}}\right) \) or \( \left( \frac{17}{40\sqrt{5}}, \frac{11}{10\sqrt{5}} \right) \).

The x coordinate of the focus of our original parabola is at \(x' = x\cos\theta - y\sin\theta = \left(\frac{17}{40\sqrt{5}}\right)\left(\frac{2}{\sqrt{5}}\right) - \left(\frac{11}{10\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}}\right) = -\frac{1}{20} \).

The y coordinate is \(y' = x\sin\theta + y\cos\theta = \left(\frac{17}{40\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}}\right) + \left(\frac{11}{10\sqrt{5}}\right)\left(\frac{2}{\sqrt{5}}\right) = \frac{21}{40} \).

The focus of the original parabola is located at \( \left(-\frac{1}{20}, \frac{21}{40}\right) \). As usual, the numbers always come out to be nice rational numbers if the coefficients are rational numbers.

The Directrix: Since we have come this far, let’s continue and find the directrix of the original parabola and prove the parabola ratio is 1. See Focus-Directrix Equation of a Parabola as we will the formula for the slope and y-intercept to find the equation of the directrix.

The slope of the directrix has an easy formula: \(m = \frac{B}{2A} = \frac{-4}{2\cdot1} = -2\).

The y-intercept formula is \(c = \frac{B^2F + 4A^2F - A(D^2 + E^2)}{2ABD - 4A^2E}\). Plugging in the numbers: \(c = \frac{(-4)^2(1) + 4(1)^2(1) - (1)((3)^2 + (-4)^2)}{2(1)(-4)(3) - 4(1)^2(-4)} = \frac{16+4-(9+16)}{-24+16} = \frac{5}{8}\). Therefore, we expect our directrix equation to be \(y = -2x + \frac{5}{8} \).

Figure 9 shows the two foci creating our angle of rotation. Also the directrix is shown in green and a point P on the parabola gives us a ratio of 1.

In the above example, we used the trick of finding the inverse, using the rotation formulas, then finding the inverse again to rotate it. Next, we will just use the second equation for |C| > |A| to rotate the parabola \(x^2 + 4xy +4y^2 + 3x - 4y + 1 = 0 \)

Example 5

Problem: Rotate the parabola \(x^2 + 4xy +4y^2 + 3x - 4y + 1 = 0 \) to standard orientation, find its vertex and focus. Then, find its directrix equation.

Solution: The angle of rotation is the same as Example 4 except it is negative of the above: \(\theta = \ \frac{1}{2}\cot^{-1}\left(\frac{1-4}{4}\right) = -26.57\). The sine and cosine are: \(\cos\theta = \frac{\sqrt{4}}{\sqrt{1+4}} = \frac{2}{\sqrt{5}}\) and \(\sin\theta = \frac{\sqrt{1}}{\sqrt{1+4}} = \frac{1}{\sqrt{5}}\) (same as Example 4).

We will use the formula: \((A+C)y^2 + \left(\dfrac{D\sqrt{|C|} \pm E\sqrt{|A|}}{\sqrt{|A+C|}}\right)x \text{ } + \) \( \left(\dfrac{E\sqrt{|C|} \mp D\sqrt{|A|}}{\sqrt{|A+C|}}\right)y + F = 0 \) because |C| > |A|.

(i) \((1+4)y^2 + \left(\dfrac{3\sqrt{4} - (-4)\sqrt{1}}{\sqrt{5}}\right)x \text{ } + \) \( \left(\dfrac{(-4)\sqrt{4} + 3\sqrt{1}}{\sqrt{5}}\right)y + 1 = 0 \)

(ii) \(5y^2 + \dfrac{10}{\sqrt{5}}x - \dfrac{5}{\sqrt{5}}y + 1 = 0 \)

Our parabola, \(5y^2 + \dfrac{10}{\sqrt{5}}x - \dfrac{5}{\sqrt{5}}y + 1 = 0 \) or \(x = -\frac{\sqrt{5}}{2}y^2+\frac{1}{2}y-\frac{\sqrt{5}}{10}\), which opens left is shown in the image below with its angle of rotation.

Vertex: The vertex is easiest to find because the formula for the y coordinate is based on the coefficients: \(-\frac{b}{2a} = \frac{-\frac{1}{2}}{2\cdot-\frac{\sqrt{5}}{2}} = \frac{1}{2\sqrt{5}}\). By substitution of the x value, the y coordinate is at: \(x = -\frac{\sqrt{5}}{2}\left(\frac{1}{2\sqrt{5}}\right)^2+\frac{1}{2}\left(\frac{1}{2\sqrt{5}}\right)-\frac{\sqrt{5}}{10} =\) \( -\frac{\sqrt{5}}{40} + \frac{1}{4\sqrt{5}} - \frac{\sqrt{5}}{10} = -\frac{3\sqrt{5}}{40} \).

The vertex of the standard parabola is at \(\left(-\dfrac{3\sqrt{5}}{40}, \dfrac{\sqrt{5}}{10} \right)\).

We will rotate these coordinates in the negative direction to find the vertex of the rotated parabola. We already found the sine and cosine of the rotation angle above. Just remember than \(\cos(-\theta) = \cos\theta\) and \(\sin(-\theta) = -\sin\theta\).

The x coordinate of the focus of our original parabola is at \(x' = x\cos(-\theta) - y\sin(-\theta) = \left(-\frac{3\sqrt{5}}{40}\right)\left(\frac{2}{\sqrt{5}}\right) + \left(\frac{\sqrt{5}}{10}\right)\left(\frac{1}{\sqrt{5}}\right) = -\frac{1}{20} \).

The y coordinate is \(y' = x\sin(-\theta) + y\cos(-\theta) = \left(\frac{3\sqrt{5}}{40}\right)\left(\frac{1}{\sqrt{5}}\right) + \left(\frac{\sqrt{5}}{10}\right)\left(\frac{2}{\sqrt{5}}\right) = \frac{11}{40} \).

The vertex of the original parabola is at \( \left(-\dfrac{1}{20}, \dfrac{11}{40}\right) \).

The Focus: The focal distance is \(\frac{1}{4a} = \frac{1}{4\cdot\frac{\sqrt{5}}{2}} = \frac{\sqrt{5}}{10}\). Since the parabola opens left, the focus of the standard parabola is to the left of the vertex and same height as the vertex. Therefore, the focus is at \(\left(-\frac{3\sqrt{5}}{40} - \frac{\sqrt{5}}{10} , \frac{\sqrt{5}}{10} \right)\) = \(\left(-\frac{3\sqrt{5}}{40} - \frac{\sqrt{5}}{10} , \frac{\sqrt{5}}{10} \right)\) = \(\left(-\frac{7\sqrt{5}}{40} , \frac{\sqrt{5}}{10} \right)\).

The focus of the standard parabola is at \(\left(-\dfrac{7\sqrt{5}}{40} , \dfrac{\sqrt{5}}{10} \right)\).

Now, we can rotate the focus to find the original parabola focus.

The x coordinate of the focus of our original parabola is at \(x' = x\cos(-\theta) - y\sin(-\theta) = \left(-\frac{7\sqrt{5}}{40}\right)\left(\frac{2}{\sqrt{5}}\right) + \left(\frac{\sqrt{5}}{10}\right)\left(\frac{1}{\sqrt{5}}\right) = -\frac{1}{4} \).

The y coordinate is \(y' = x\sin(-\theta) + y\cos(-\theta) = \left(\frac{7\sqrt{5}}{40}\right)\left(\frac{1}{\sqrt{5}}\right) + \left(\frac{\sqrt{5}}{10}\right)\left(\frac{2}{\sqrt{5}}\right) = \frac{3}{8} \).

The focus of the original parabola is at \( \left(-\dfrac{1}{4}, \dfrac{3}{8}\right) \). Figure 12 shows our vertices and foci of both parabolas, with focal distances shown to be equal to \(\frac{\sqrt{5}}{10}\) and location of the focus of the original parabola.

The Directrix: Lastly, we want to find the directrix of the original parabola. See Focus-Directrix Equation of a Parabola for the formulas we will use here.

The slope of the directrix is quite easy because its formula is \(\frac{B}{2A}\). The parabola equation is \(x^2+4xy+4y^2+3x-4y+1=0\) So the slope of this parabola is \(\frac{4}{2\cdot1} = 2\).

The y-intercept of the directrix is given by the equation \( \frac{4CF + 4AF - D^2 - E^2}{2BD - 4AE}\). Plugging in the numbers: \( \frac{4(4)(1) + 4(1)(1) - (3)^2 - (-4)^2}{2(4)(3) - 4(1)(-4)} = -\frac{1}{8}\).

Therefore, the equation of the directrix of the original parabola is \(y = 2x - \frac{1}{8}\). The image below shows a zoomed out image with the parabola ratio equal to 1.

Rotating Parabolas from Standard Position

Now, let’s rotate a parabola from standard orientation to any angle θ. We will use the equation of the parabola \(y = ax^2 + bx + c\) to see where a, b, and c end up.

To rotate counterclockwise, we use the following substitutions: \(x' = x\cos\theta + y\sin\theta; \text{ } y' = y\cos\theta - x\sin\theta\)

(i) \(ax^2 - y + bx + c = 0\)

(ii) \(a(x\cos\theta + y\sin\theta)^2 - (y\cos\theta - x\sin\theta) +\) \( b(x\cos\theta + y\sin\theta) + c = 0\)

A parabola rotated by an angle θ counterclockwise has the equation: \((a\cos^{2}\theta)x^2 + a\sin(2\theta)xy + \) \( (a\sin^{2}\theta)y^2 + (b\cos\theta + \sin\theta)x + (b\sin\theta - \cos\theta)y \) \( + c = 0\)

Example

Let’s rotate the parabola \(\frac{1}{2}x^2 - y - 4x + 2 = 0\) by 30 degrees or π/6.

First, we will evaluate the sine and cosine values needed: \(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\), \(\sin\frac{\pi}{6} = \frac{1}{2}\), and \(\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\). Applying the values:

(i) \(\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)^{2}x^2 + \frac{1}{2}\cdot\frac{\sqrt{3}}{2}xy + \frac{1}{2}\left(\frac{1}{2}\right)^{2}y^2 + (-4\cdot\frac{\sqrt{3}}{2} + \frac{1}{2})x + (-4\cdot\frac{1}{2} - \frac{\sqrt{3}}{2})y + 2 = 0\)

(ii) \(\frac{3}{8}x^2 + \frac{\sqrt{3}}{4}xy + \frac{1}{8}y^2 + \frac{1}{2}(1 - 4\sqrt{3})x - \frac{1}{2}(4 + \sqrt{3})y + 2 = 0\)

Notice that \(A+C = \frac{1}{8} + \frac{3}{8} = \frac{1}{2}\), which is equal to a, our original coefficient. In our trigonometric form, the values of A and C are \(a\cos^{2}\theta\) and \(a\sin^{2}\theta\), so they do add to a.

Figure 2 shows our result. The vertex has been identified and a tangent has been drawn. The angle between the vertices of the parabolas with respect to the origin is shown and the distance to the two vertices and the origin is also shown.

The Vertex

Remember not to confuse the uppercase coefficients used for general conics with the lowercase coefficients used specifically for parabolas.

Rotation of the Parabola Activity

In this Geogebra interaction, you can modify the coefficients a, b, and c of the parabola, and then use the slider to change the angle of rotation to rotate the parabola.

The activity shows the focus of the parabola and the vertex. The formulas used for these two points were ones we derived here.

The rotation occurs counterclockwise for positive angle measures and clockwise for negative angle measures.

The General Parabola Equation

Equations of the form \(\sqrt{ax} + \sqrt{by} = \sqrt{c}\) are parabolas. When squaring twice, the general form is \(a^2x^2 - 2abxy + b^2y^2 - 2acx - 2bcy + c^2 = 0\). The determinant of this equation is \((-2ab)^2 - 4a^2b^2 = 0\). Therefore, this equation type is always a parabola.

We can rewrite the above using our capital coefficients. Because c can be any number, the x and y coefficients can be unique and do not affect the coefficients of x² and y². Therefore, we can remove the variable c and write our equation as \(A^2x^2 - 2ABxy + B^2y^2 + Ex + Dy + F = 0\).

The Geogebra activity below allows you to play with the coefficients and observe that it is a parabola for any value.

Parametric Equation of Parabola

The parabola does not have a parametric form in terms of trigonometric functions like the other 3 conics. Therefore, we will just write our rectangular as a parametric equation the easy way:

\(x(t) = t\)  and  \(y(t) = at^2\)

We can rotate this form by using the rotation formulas.

A parabola rotated by θ° is given by: \(x'(t) = t\cos\theta + at^2\sin\theta\)  and  \(y'(t) = at^2\cos\theta - t\sin\theta\)

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