# Tangents of an Hyperbola

Just like an ellipse, the hyperbola’s tangent can be defined by the slope, m, and the length of the major and minor axes, without having to know the coordinates of the point of tangency.

A tangent to a hyperbola $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$ with a slope of m has the equation $$y = mx \pm \sqrt{a^2m^2 - b^2}$$.

(1) There are two tangents to the hyperbola with the same slope of m. Both tangents will be parellel. And of course, a chord connecting the two tangent points will pass through the center of the hyperbola because the points are opposite of each other.

(2) The equation of the tangent can be stated completely in terms of the slope, m, and the lengths of the major and minor axes, a and b, without having to determine the point of tangency. And this is a profound statement, which lead to the next point.

(3) It follows from (1) and (2) that there are two y-intercepts that are determined by the slope of the hyperbola. In other words, if we have determined m, then the y-intercept can be stated by m, a, and b, reglardless of the coordinates of the point of tangency.

(4) Unlike an ellipse, which covers all slopes possible from negative to infinity to infinity, the hyperbola does not cover all the slopes possible. The slope is bound by the asymptotes of the hyperbola.

Figure 1 shows two parallel tangents to a hyperbola at A and D that have the same slope. There is no other tangent line except these two that can have this slope. These two tangents have orthogonals at B and C that create a right angle at I1, I2, I3, and I4. These points lie on a circle called the director circle.

## The Equation of the Tangent

Let’s first find the equation of the tangent in the slope-intercept form. Since we have determined our slope will be m, we simply need to find the y-intercept. The y-intercept of a tangent to a hyperbola at (x0, y0) is given by $$\dfrac{b^2}{y_0}$$. The derivation of this formula has been covered on the Conic Sections page.

The equation of the tangent line to a hyperbola $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$ with slope m is $$y = mx - \dfrac{b^2}{y_0}$$.

So far, it seems we need to know the y coordinate of the point of tangency to determine the equation of the line, which contradicts statement (2) above. But there is a relationship of y0 with the slope. It is actually straight-forward. Let’s first find the derivative of our hyperbola. We will solve for y to find the derivative.

(i) $$y = \pm \dfrac{b}{a}\sqrt{x^2-a^2}$$

(ii) $$\dfrac{dy}{dx} = \pm \dfrac{b}{a}\cdot\dfrac{2x}{2\sqrt{x^2 - a^2}}$$

(iii) $$\dfrac{dy}{dx} = \pm \dfrac{bx}{a\sqrt{x^2-a^2}}$$

Now, we will substitute x0 for x to find the slope at this point. This point will be the point at which our slope will be m. Therefore,

(iv) $$m = \pm \dfrac{bx_0}{a\sqrt{x^2-a_{0}^2}}$$

In equation (iv), we can solve for x0 to find its location in terms of a, b, and m!

(v) $$m^2 = \dfrac{b^2x^2}{a^2(x^2-a_{0}^2)}$$

(vi) $$x_0 = \dfrac{a^2m}{\pm\sqrt{a^2m^2-b^2}}$$ (Rearranging to solve for x0.)

We have solved for x0 in terms of the slope and the length of major and minor axes. These two x coordinates are unique for the slope. Now, to find the y0, we simply substitute this value in our hyperbola equation and solve for y0.

(vii) $$\dfrac{\dfrac{a^4 m^2}{a^2m^2-b^2}}{a^2} + \dfrac{y_0^2}{b^2} = 1$$

(viii) $$\dfrac{\dfrac{a^2 m^2}{a^2m^2-b^2}}{a^2} + \dfrac{y_0^2}{b^2} = 1$$

(ix) $$y_0^2 = \dfrac{b^4+b^2a^2m^2-b^2a^2m^2}{a^2m^2-b^2}$$ (Skipping some elementary algebra steps in between.)

(x) $$y_0 = \dfrac{b^2}{\pm\sqrt{a^2m^2+b^2}}$$

And now we have our y coordinate in terms of a, b, and m! We can now substitute this value in our previous equation of the tangent line:

(xi) $$y = mx + \dfrac{b^2}{\dfrac{b^2}{\pm\sqrt{a^2m^2-b^2}}}$$

(xii) $$y = mx \pm\sqrt{a^2m^2-b^2}$$.

Finally, we have our elusive equation of the tangent to a hyperbola completly given by its slope and the lengths of the major and minor axes.

## The Director Circle of the Hyperbola

The above equation for the tangent line allows us to find the equation of the director circle of the hyperbola.

The director circle is the locus of all points of intersection of orthogonal tangents to a hyperbola. In other words, the intersection points of two tangents that form a right angle fall on a circle. Figure 1 shows the director circle in red.

The director circle of the hyperbola $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$ is given by the equation $$x^2+y^2=a^2-b^2$$.

### Proof of the Director Circle Equation

A tangent with slope m has an orthogonal with slope –1/m. Therefore, our pair of orthogonals is: $$y = mx \pm \sqrt{a^2m^2-b^2}$$ and $$y = -\dfrac{1}{m}x \pm \sqrt{a^2\left(-\dfrac{1}{m}\right)^2-b^2}$$. Our goal is to eliminate m and find the resulting equation based totally on x and y and any other variables (i.e. a and b).

Let’s isolate the radicals for both equations and square both sides:

(i) $$(y - mx)^2 = (\pm\sqrt{a^2m^2-b^2})^2$$;    $$\left(y +\dfrac{1}{m}x\right)^2 = \left(\pm \sqrt{a^2\left(-\dfrac{1}{m}\right)^2-b^2}\right)^2$$

(ii) $$y^2-2mxy +m^2x^2 = a^2m^2-b^2$$;    $$y^2+\dfrac{2}{m}xy+\dfrac{1}{m^2}x^2=\dfrac{a^2}{m^2}-b^2$$

We will multiply the second tangent equation by m² to eliminate the denominators.

(iii) $$y^2-2mxy + m^2x^2 = a^2m^2-b^2$$;    $$m^2y^2+2mxy+x^2=a^2-m^2b^2$$

Now, we will add the two equations to eliminate the xy term.

(iv) $$(1+m^2)y^2+(1+m^2)x^2 = (1+m^2)a^2 - (1+m^2)b^2$$.

(v) $$x^2 + y^2 = a^2 - b^2$$.

The (1 + m²) conveniently canceled out leaving us with the equation of the director circle that has the radius of $$\sqrt{a^2-b^2}$$.

The director circle of a hyperbola $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2} = 1$$ is the locus of all points of intersection of two tangents to a hyperbola that create a right angle. The equation of the director circle is $$x^2 + y^2 = a^2 - b^2$$.

## Alternate Equation of Hyperbola Tangents

There is another equation for the tangents to a hyperbola that does not involve the slope of the line. For example, if one does not know the slope but knows the coordinates of the point of tangency to a hyperbola, then this equation is better suited.

The equation of a tangent to a hyperbola $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2} = 1$$ at point (x0, y0) is given by: $$\dfrac{x_0}{a^2}x - \dfrac{y_0}{b^2}y = 1$$.

Note how similar the tangent equation is to the hyperbola equation.

### Proof

Deriving this equation is a matter of rearranging the terms. Remember the slope of a tangent at (x0, y0) is given by $$\dfrac{b^2 x_0}{a^2 y_0}$$. Since we have the slope and a point, we can use the point-slope form of the line equation and manipulate the equation.

(i) $$y - y_0 = \dfrac{b^2 x_0}{a^2 y_0}(x-x_0)$$

(ii) $$y = \dfrac{b^2 x_0}{a^2 y_0}x - \dfrac{b^2 x_0^2}{a^2 y_0} + y_0$$

(iii) $$y = \dfrac{b^2 x_0}{a^2 y_0}x - \dfrac{b^2 x_0^2 - a^2 y_0^2}{a^2 y_0}$$

The quantity $$b^2 x_0^2 - a^2 y_0^2$$ is equal to $$a^2b^2$$ because this is a point on the hyperbola (after rewriting the hyperbola equation without denominators).

(iv) $$y = \dfrac{b^2 x_0}{a^2 y_0}x - \dfrac{a^2b^2}{a^2 y_0}$$

(v) $$y = \dfrac{b^2 x_0}{a^2 y_0}x - \dfrac{b^2}{y_0}$$

Well, we already knew the y-intercept was $$\dfrac{b^2}{y_0}$$. We will multiply both sides by $$-\dfrac{y_0}{b^2}$$.

(vi) $$-\dfrac{y_0}{b^2}y = -\dfrac{x_0}{a^2}x + 1$$

Rearranging (vi) gives us the equation we needed to find for our tangent.

## Bounds for Tangent Lines to a Hyperbola

Any point on an ellipse can have orthogonals. This is not the case for hyperbolas.

If the hyperbola’s asymptotes create an angle that is greater than or equal to 90°, it will not have orthogonals. That means if $$b^2 = a^2$$, then the director circle degenerates into a point as show in in Figure 2. That means the asymptotes create a right angle at their intersection.

And if $$b^2 > a^2$$, then the director circle no longer exists since the radius is negative as shown in Figure 3. Therefore, there can be no orthogonal tangents on the same branch if the asymptotes create an angle greater than 90° because all the angle created by the intersection are obtuse and angles created by points on the two different branches are all acute.

In Figure 1 (in which $$a^2 > b^2$$), as point A moves toward the vertex, then at some point, there is no point B on the same branch which will create an orthogonal. The point at which that happens is when the tangent at A is also tangent to the director circle.

Figure 4 shows that when tangent at A is also tangent at E, point B on the same branch of the hyperbola no longer exists which can create a right angle at E. Actually, point B disappeared into negative infinity and the tangent at B became asymptote of the hyperbola - which does form a right angle at E.

Therefore, we expect that happens when the slope of the tangent is $$\dfrac{b}{a}$$. We will prove just this.

### Proof

The equation of a tangent line to the circle is $$y = mx \pm r\sqrt{m^2+1}$$ and the hyperbola is $$y = mx \pm \sqrt{m^2a^2-b^2}$$. A common tangent to both would have the same slope and y-intercept. Let’s equate the y-intercepts of the two and find the slope.

(i) $$r\sqrt{m^2+1}\ = \sqrt{m^2a^2-b^2}$$

Since this is the director circle of the hyperbola, the radius of the director circle is $$\sqrt{a^2 - b^2}$$.

(ii) $$\sqrt{a^2 - b^2}\sqrt{m^2+1}\ = \sqrt{m^2a^2-b^2}$$

(iii) $$(a^2 - b^2)(m^2+1)\ = m^2a^2-b^2$$

(iv) $$m^2a^2 + a^2 - m^2b^2 - b^2\ = m^2a^2-b^2$$

(v) $$m^2b^2 \ = a^2$$

(vi) $$m = \pm\dfrac{a}{b}$$

There are two lines that are tangent to both the hyperola and the director circle, one on each branch of the hyperbola.

When substituting the slope to find the tangent line equation, we get $$y=\dfrac{a}{b}x \pm \dfrac{\sqrt{a^4-b^4}}{b}$$.

To find the actual point of tangency on the circle, it is easier to solve for the intersection of the director circle and the asymptotes of the hyperbola because the asymptote equation is a lot simpler: $$y = \pm\dfrac{b}{a}x$$. The two points of tangency on the circle are $$\left(\pm a\sqrt{\dfrac{a^2-b^2}{a^2+b^2}}, \mp b\sqrt{\dfrac{a^2-b^2}{a^2+b^2}}\right)$$.

The points on the hyperbola of the common tangent are: $$\left( \pm\dfrac{a^3\sqrt{a^4-b^4}}{a^4-b^4}, \pm\dfrac{b^3\sqrt{a^4-b^4}}{a^4-b^4} \right)$$.

## Last Word

Do other conics have a director circle? The ellipse does have a director circle. Its equation is $$x^2 + y^2 = a^2 + b^2$$.

For parabolas, the orthogonals lie on a straight line, specifically the directrix. The proof is given here: Properties of Parabolas.