# “Hyperbola” Functions

There are 4 conic sections: circles, ellipses, parabolas, and hyperbolas. Two of them can be represented as functions without having to use two equations to complete the curve: parabolas and hyperbolas. Circles and ellipses cannot be represented as functions without having to use two equations.

The case of parabolas is trivial because a parabola can be defined as $$y = ax^2 + bx + c$$, which are functions. Only if the parabola opens left or right can it not be represented as functions.

Similarly, hyperbolas can be rotated so that one of its asymptotes is vertical, making the curve a function. You may have already come across hyperbolas as functions. The simplest hyperbolas are $$y=\dfrac{k^2}{x}$$ or $$xy = k^2$$ for some constant k. This hyperbola is a constant product curve. The determinant is $$1^2 - 4(0)(0) = 1$$, which means this is a hyperbola.

## The Constant Product Curve

Let’s rotate the constant product curve and reveal its equation in standard form. Then, we can compare the graphs. We will use the formulas $$x' = x\cos\theta - y\sin\theta$$ and $$y' = x\sin\theta + y\cos\theta$$. The angle of rotation is 45° because the inverse of this curve is itself making the line $$y = x$$ the line of symmetry. The trigonometric values are $$\cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2}$$.

(i) $$xy = k^2$$

(ii) $$\left(\dfrac{\sqrt{2}}{2}x - \dfrac{\sqrt{2}}{2}y\right)\left(\dfrac{\sqrt{2}}{2}x + \dfrac{\sqrt{2}}{2}y \right) = k^2$$

(iii) $$\dfrac{1}{2}x^2 + \dfrac{1}{2}xy - \dfrac{1}{2}xy - \dfrac{1}{2}y^2 = k^2$$

(iv) $$\dfrac{1}{2}x^2 - \dfrac{1}{2}y^2 = k^2$$

(v) $$\dfrac{x^2}{2k^2} - \dfrac{y^2}{2k^2} = 1$$

Equation (v) puts our curve in standard form and we can see it is a hyperbola. These curves can be rotated by 45° clockwise or counterclockwise. We expected hyperbolas of this type to be a “square” hyperbola or a “right” hyperbola, meaning a and b are equal and the asymptotes are perpendicular.

The value of c is $$c = \sqrt{2k^2 + 2k^2} = 2k$$. That places the foci of the standard hyperbola at $$(\pm 2k, 0)$$. The vertices are at $$(\pm\sqrt{2}k, 0)$$ for the standard hyperbola. Since the angle of rotation is 45°, dividing by $$\sqrt{2}$$ places the foci of the original hyperbola at $$(\pm 2k, \pm 2k)$$ and the vertices at $$(\pm k, \pm k)$$. The image below shows a hyperbola with k = 2. All the critical points and the asymptotes have been shown.

## The General Hyperbola Function

The general equation of a hyperbola function is $$y = \dfrac{Ax^2+Dx+F}{Bx+E}$$. When written in standard form, we get $$Ax^2 - Bxy +Dx-Ey+F=0$$. However, ignoring the negative signs to make the equation simpler, we can write the general hyperbola function as $$Ax^2 + Bxy + Dx+Ey+F=0$$.

If the y² term is missing (C = 0), the determinant will be $$B^2-4(A)(0) = B^2$$, which will always be greater than 0. Therefore, this equation is always a hyperbola.

### The Angle of Rotation

The angle of rotation is $$\theta = \frac{1}{2}\cot^{-1}\frac{A}{B}$$. The formula is a little simpler because C = 0.

Whether the angle of rotation is positive or negative depends on the signs of A and B. However, we know that the angle of rotation is never greater than 45°. Therefore, if the line of symmetry creates an angle greater than 45° with the x-axis, the angle of rotation will be negative and we will have a hyperbola that opens up and down. If the line of symmetry creates an angle less than or equal to 45° with the x-axis, the hyperbola will open left and right.

When written in this form: $$y = \dfrac{Ax^2+Dx+F}{Bx+E}$$, if A and B are both positive or both negative, the angle of rotation will always be negative. That means the hyperbola will rotate counterclockwise, or the axis of rotation will rotate clockwise. The branches will open up or down.

And if the signs are opposite, the hyperbola will always rotate clockwise and the branches will open left and right.

### The Coefficients

Now, we are ready for the coefficients:

Let’s determine the coefficients of this hyperbola when rotated to standard orientation. Unfortunately, since B may be positive or negative, the angle of rotation can be either. So our coefficients will have some conditionals as mentioned in Rotation of Conics.

(i) $$A' = \dfrac{A}{2} + \dfrac{A|A|\pm |B|\cdot B}{2\sqrt{A^2+B^2}}$$

(ii) $$C' = \dfrac{A}{2} - \dfrac{A|A|\pm |B|\cdot B}{2\sqrt{A^2+B^2}}$$

(iii) $$D' = D\sqrt{\dfrac{1}{2}+\dfrac{|A|}{2\sqrt{A^2+B^2}}} \text{ }\pm$$ $$E\sqrt{\dfrac{1}{2}-\dfrac{|A|}{2\sqrt{A^2+B^2}}}$$.

(iv) $$E' = E\sqrt{\dfrac{1}{2}+\dfrac{|A|}{2\sqrt{A^2+B^2}}}\text{ } \mp$$ $$D\sqrt{\dfrac{1}{2}-\dfrac{|A|}{2\sqrt{A^2+B^2}}}$$.

(vi) $$F' = F$$

If the angle of rotation is positive, we use the top sign and if the angle or rotation is negative, we use the bottom sign. Unfortunately, we cannot combine $$A|A|$$ as $$A^2$$ because the values came from the trigonometric evaluation.

It is best to do an example of rotating a hyperbola.

### Example 1

Problem: Rotate the hyperbola $$y= \dfrac{2x^2-4x+5}{x-6}$$ to standard orientation. Find its asymptotes, the foci and vertices.

Slant asymptote: Let’s start with the asymptotes first since we can find those just from the equation above. We can divide to find the slanted asymptote: $$y = 2x + 8 + \dfrac{53}{x-6}$$. Therefore, the slant asymptote is $$y = 2x+8$$.

Vertical asymptote: Vertical asymptote is where the denominator is undefined or equal to 0. Therefore, $$x = 6$$ is the vertical asymptote.

The Center: The center of the rotated hyperbola is at the intersection of the two asymptotes that we just found. The x coordinate is 6. The y coordinate can be found by substituting this value to the slant asymptote equation: $$y = 2(6)+8 = 20$$. So the center is (6,20).

To find the center of the standard orientation hyperbola, we can complete the square as we have done later or use the rotation formulas by utilizing the sine and cosine of the rotation angle that we will find next.

Angle of rotation: The angle is given by $$\theta = \dfrac{1}{2}\cot^{-1}\dfrac{2}{-1} = -13.28^{\circ}$$. We just need to know if it’s positive or negative to apply the correct sign. Since it is negative, we use the bottom sign.

Let’s calculate the trigonometric values because they will be necessary to find the foci. The sides of our right triangle will be 2 and 1 (equal to |A| and |B|). The hypotenuse will be $$\sqrt{5}$$. Using the half-angle identities: $$\cos\theta = \sqrt{\frac{1}{2}+\frac{1}{2}\cdot\frac{2}{\sqrt{5}}} = \sqrt{\frac{5+2\sqrt{5}}{10}}$$ and $$\sin\theta = \sqrt{\frac{5-2\sqrt{5}}{10}}$$.

The Coefficients:

We need to write our equation in general form to evalue the coefficients. Our rational equation in general form is $$2x^2-xy-4x+6y+5=0$$. Now, we can easily spot the coefficients of the rotated hyperbola.

(i) $$A' = \dfrac{2}{2} + \dfrac{2|2| - |1|\cdot (-1)}{2\sqrt{2^2+1^2}} = \dfrac{2+\sqrt{5}}{2}$$

(i) $$C' = \dfrac{2}{2} - \dfrac{2|2| - |1|\cdot (-1)}{2\sqrt{2^2+1^2}} = \dfrac{2-\sqrt{5}}{2}$$

(iii) $$D' = -4\sqrt{\dfrac{1}{2}+\dfrac{|2|}{2\sqrt{2^2+(-1)^2}}} -$$ $$6\sqrt{\dfrac{1}{2}-\dfrac{|2|}{2\sqrt{2^2+(-1)^2}}} =$$ $$-4\sqrt{\dfrac{5 + 2\sqrt{5}}{10}} -6\sqrt{\dfrac{5 - 2\sqrt{5}}{10}} =$$ $$-\sqrt{\dfrac{130+4\sqrt{5}}{5}}$$.

(iv) $$E' = 6\sqrt{\dfrac{5 + 2\sqrt{5}}{10}} - 4\sqrt{\dfrac{5 - 2\sqrt{5}}{10}} =$$ $$\sqrt{\dfrac{130-4\sqrt{5}}{5}}$$

(vi) $$F' = 5$$

So far, our equation for the standard orientation hyperbola is $$\left(\frac{2+\sqrt{5}}{2}\right)x^2 + \left(\frac{2-\sqrt{5}}{2}\right)y^2 - \left(\sqrt{\frac{130+4\sqrt{5}}{5}}\right)x + \left(\sqrt{\frac{130-4\sqrt{5}}{5}}\right)y + 5 = 0$$.

Now, we need to complete the square, which can be tedious. We can use Wolfram to help simplify our radical expressions.

(i) $$\left(\dfrac{\sqrt{5}+2}{2}\right)\left( x - \sqrt{\dfrac{1090-484\sqrt{5}}{5}} \right)^2 - \dfrac{61\sqrt{5}-120}{5} -$$ $$\left(\dfrac{\sqrt{5} - 2}{2}\right)\left( y - \sqrt{\dfrac{1090+484\sqrt{5}}{5}} \right)^2 + \dfrac{61\sqrt{5}+120}{5} + 5 = 0$$

(ii) $$\left(\dfrac{2+\sqrt{5}}{2}\right)\left( x - \sqrt{\dfrac{1090-484\sqrt{5}}{5}} \right)^2 -$$ $$\left(\dfrac{\sqrt{5}-2}{2}\right)\left( y - \sqrt{\dfrac{1090+484\sqrt{5}}{5}} \right)^2 = -53$$

(iii) $$-\dfrac{\left( x - \sqrt{\dfrac{1090-484\sqrt{5}}{5}} \right)^2}{\dfrac{3690}{\sqrt{5}+2}} +$$ $$\dfrac{\left( y - \sqrt{\dfrac{1090+484\sqrt{5}}{5}} \right)^2}{\dfrac{3690}{\sqrt{5}-2}} = 1$$

(iv) $$-\dfrac{\left( x - \sqrt{\dfrac{1090-484\sqrt{5}}{5}} \right)^2}{106\sqrt{5}-212} +$$ $$\dfrac{\left( y - \sqrt{\dfrac{1090+484\sqrt{5}}{5}} \right)^2}{106\sqrt{5}+212} = 1$$

The minor axis (2b) is about 10.0 and the major axis (2a) is about 42.38. There’s quite a difference so the hyperbola will be narrow.

I wrestled with this all day to complete the square! Not a pretty equation.

Asymptotes: The slope of the asymptotes of a hyperbola that has a vertical transverse axis is $$\pm\dfrac{a}{b}$$, where a is the constant of the y² term and b is the constant of the x² term. The slope of the above hyperbola is $$\pm\frac{\sqrt{106\sqrt{5}+212}}{\sqrt{106\sqrt{5}-212}} = \pm \sqrt{\frac{\sqrt{5}+2}{\sqrt{5}-2}} = \pm\sqrt{9+4\sqrt{5}}$$. Well, at least the slope simpified nicely.

We need to translate the line with this slope based on the center of the hyperbola. So the two asymptotes are located at $$y = \left(\pm\sqrt{9+4\sqrt{5}}\right)\left(x - \sqrt{\frac{1090-484\sqrt{5}}{5}}\right) + \sqrt{\frac{1090+484\sqrt{5}}{5}}$$ or $$y = \left(\pm\sqrt{9+4\sqrt{5}}\right)x \pm \sqrt{\frac{130-4\sqrt{5}}{5}} + \sqrt{\frac{1090+484\sqrt{5}}{5}}$$ or

(1) $$y = \left(\sqrt{9+4\sqrt{5}}\right)x -8\sqrt{\frac{2+4\sqrt{5}}{5}}$$ and

(2) $$y = \left(-\sqrt{9+4\sqrt{5}}\right)x + 6\sqrt{10+4\sqrt{5}}$$

are the two asymptotes.

Foci and eccentricity: Both of these depend on c, so we can calculate that first: $$c = \sqrt{106\sqrt{5} - 212 + 106\sqrt{5} + 212} = 2\sqrt{53\sqrt{5}}$$. The eccentricity is $$e = \frac{\sqrt{106\sqrt{5}}}{\sqrt{106\sqrt{5}-212}} = \sqrt{\frac{\sqrt{5}}{\sqrt{5}-2}} = \sqrt{5+2\sqrt{5}} \approx 3.078$$.

To find the foci, we use the focal distance, c, as the hypotenuse of the right triangle. The horizontal and vertical displacements are reversed because the triangle is oriented down. The horizontal displacement is $$\Delta x = c\sin\theta = 2\sqrt{53\sqrt{5}}\cdot\sqrt{\frac{5-2\sqrt{5}}{10}} = 2\sqrt{\frac{53\sqrt{5}-160}{2}}$$ and $$\Delta y = c\cos\theta = 2\sqrt{53\sqrt{5}}\cdot\sqrt{\frac{5+2\sqrt{5}}{10}} = 2\sqrt{\frac{53\sqrt{5}+160}{2}}$$. These displacements are from the center of the hyperbola. So the foci are at $$\left(6+2\sqrt{\frac{53\sqrt{5}-160}{2}} ,20+2\sqrt{\frac{53\sqrt{5}+160}{2}}\right)$$.

Figures 3 show a zoom in of the focus because it is very close to the vertex and a zoom out because the distance between the foci is large. The subtraction ratio of the hyperbola is about 42.38 as calculated by Geogebra. This should equal the major axis length, which is twice a: $$2\sqrt{106\sqrt{5}+212} = 42.38$$. It is exact by the hundredth, so we are certain our calculations are correct.

Transverse axis: We can find the equation of the transverse axis of the slanted hyperbola. The angle of the transverse axis is the average of the angles of the two asymptotes. Let φ be the angle of the slanted asymptote. Then, the angle of the transverse axis is $$\dfrac{\frac{\pi}{2}+\phi}{2}$$ or $$\dfrac{\pi}{4}+\dfrac{\phi}{2}$$. The tangent of this is the slope of the transverse axis. Therefore, $$m = \tan\left(\dfrac{\pi}{4}+\dfrac{\phi}{2}\right)$$. Using the sum property of tangent: $$m = \dfrac{1+\tan\frac{\phi}{2}}{1-\tan\frac{\phi}{2}}$$. We already know the slope of the slant asymptote is 2. Therefore, $$\tan\phi = 2$$. Using the identity $$\tan\phi = \dfrac{2\tan\left(\frac{1}{2}\phi\right)}{1-\tan^{2}\left(\frac{1}{2}\phi\right)}$$, we can solve for $$\tan\frac{\phi}{2}$$ (by replacing $$\tan\frac{\phi}{2}$$ with x and solving a quadratic equation): $$\tan\frac{\phi}{2} = \frac{\sqrt{5}-1}{2}$$ (we rejected the negative value because the slope is positive). The slope of the transverse axis is: $$m = \dfrac{1+ \frac{\sqrt{5}-1}{2}}{1-\frac{\sqrt{5}-1}{2}} = \sqrt{5} + 2$$.

The intersection point of the two asymptotes is our translation values for the line. The two asymptotes intersect at (6,20). Therefore, our equation for the transverse axis is $$y - 20 = (\sqrt{5} + 2)(x-6)$$ or $$y = (\sqrt{5} + 2)x - 6\sqrt{5}+8$$.