Conic SectionsTangents of an Ellipse
Browsing through Google searches about tangents to an ellipse, I found an equation that I had never seen before, and none of the sites mention how to prove this relationship. Determined to find the proof, I finally found it, and it was not at all hard. I will go through the proof of the equation of a tangent to an ellipse, while noting some properties.
A tangent to an ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) with a slope of m has the equation \(y = mx \pm \sqrt{a^2m^2 + b^2}\).
What is so big deal about the above equation anyway? Well, it reveals a few properties of ellipses (and circles).
(1) There are two tangents to the ellipse 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 ellipse 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 ellipse. 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) The ellipse covers all slopes possible from negative to infinity to infinity. It takes a quarter of the ellipse to cover all the negative slopes and another quarter to cover all the positive slopes. There are exactly 2 unique lines with the same slope; therefore, the slope can be determined without knowing the point of tangency.
The figure above shows two parallel tangents to an ellipse that have the same slope. There is no other tangent line except these two that can have this slope. There are no other tangents to this ellipse that will have the slope depicted in Figure 1.
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 an ellipse 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 an ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) 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. This is where I spent quite some time finding the relationship of y0 with the slope. It is actually straight-forward. Let’s first find the derivative of our ellipse. We will solve for y to find the derivative.
(i) \(y = \pm \dfrac{b}{a}\sqrt{a^2-x^2}\)
(ii) \(\dfrac{dy}{dx} = \pm \dfrac{b}{a}\cdot\dfrac{-2x}{2\sqrt{a^2-x^2}}\)
(iii) \(\dfrac{dy}{dx} = \pm \dfrac{bx}{a\sqrt{a^2-x^2}}\)
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) \(m = \pm \dfrac{bx_0}{a\sqrt{a^2-x_{0}^2}}\)
In equation (iv), we can solve for x0 to find its location in terms of a, b, and m!
(v) \(m^2 = \dfrac{b^2x^2}{a^2(a^2-x_{0}^2)}\)
(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 ellipse 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 an ellipse completly given by its slope and the lengths of the major and minor axes.
The Director Circle
The above equation for the tangent line allows us to find the equation of the director circle of the ellipse. Why is it called the director circle anyway?
The director circle is the locus of all points of intersection of orthogonal tangents to an ellipse. In other words, the intersection points of two tangents that form a right angle fall on a circle. This has been covered here briefly (without proof) on the Conic Sections page. The image below shows a director circle for an ellipse.
In Figure 2, Point L is an arbitrary point that creates a right angle to two tangents. Point P is the intersection of the horizontal and vertical tangents of the ellipse. As one may deduce from point P, the radius of the circle must be the hypotenuse of the half the major and minor axes, or \(\sqrt{a^2+b^2}\). We can prove this using our tangent line equations, however.
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) \(x^2 + y^2 = a^2 + b^2\).
How convenient that (1 + m²) 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 an ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1\) is the locus of all points of intersection of two tangents to an ellipse that create a right angle. The equation of the director circle is \(x^2 + y^2 = a^2 + b^2\).
Since a = b = r in a circle, the equation of the director circle of a circle is given below.
The director circle of a circle \(x^2 + y^2 = r^2\) is the locus of all points of intersection of two tangents to a that create a right angle. The equation of the director circle is \(x^2 + y^2 = 2r^2\).
Alternate Equation of Ellipse Tangents
There is another equation for the tangents to an ellipse that does not involve the slope of the line. For example, if one does not know the slope but knows the coordinates of the ellipse, then this equation is better suited.
The equation of a tangent to an ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1\) 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 ellipse 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 ellipse (after rewriting the ellipse 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.
Last Word
Do other conics have a director circle? The hyperbola does have a director circle. Its equation is \(x^2 + y^2 = a^2 - b^2\). It is not possible for all hyperbolas to have orthogonals. Particularly, if a hyperbola’s asymptotes create an angle greater than 90° on the side of the two branches, then it will not have orthogonal tangents.
For parabolas, the orthogonals lie on a straight line, specifically the directrix. The proof is given here: Properties of Parabolas.