Chapter 4.4 Symmetry of Tangential Intersections

Tangential Intersection

We investigate special kind of intersections that have been studied extensively for over 2000 years, from as early as Euclid.

Given a point $p$, it is always possible to construct a line that intersect the point. For example, if $p=(1,-4)$, then the line $L$ given by \[L(x)=3(x-1)-4\] will intersect at the point $p$:

In this example, the line has a slope of $m=3$, but any slope can work.

Now what about intersection of a line and a function? For example, take $f(x)=x^2-1.$ Is there a line that intersects $f$ at $x=2$? The answer is yes. We just need a line that passes through the point $(2,f(2))$, which in this case is the point $(2,3)$. Here are three examples of lines that intersect $f$ at $x=2.$

Each line intersects $f$ at $(2,3)$ and each has a different slope. Notice that in Example A and Example B, the lines intersect not just at $(2,3)$, but at another point as well. These are known as transversal intersections. However, Example C only intersects once. This is known as a tangential intersection. Another property of this type of intersection is that if we restrict ourselves to small enough interval that contains the $x$ coordinate of the intersection, the line and quadratic function are hardly distinguishable from one another.

Our goal is to figure out when these type of intersections will occur. We will eventually study this idea of tangential intersection in Chapter 5 and Chapter 6 in more general settings, but for now we will focus on three specific settings.

The intersection of a line and a circle
The intersection of a line and an ellipse
The intersection of a line and a polynomial.

We begin with circles and then generalize to ellipses.

Tangential Intersection of Circles and Ellipses

To study tangential intersections on circles, and then ellipses, we first study the unit circle as motivation.

Given a point $p$ on the unit circle, we want to plot lines that intersect the unit circle $p$ and only at $p$. Draw such lines at each of the points below \[(1,0),\quad (0,1),\quad (-1,0),\quad (0,-1).\]

Below are graphs of such lines at each of the points:

In these examples, they appear to be either vertical or horizontal lines. Is that always the case? What would the line have to look like at the point $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)?$

This is neither a vertical nor a horizontal line, so the observations from before don’t generalize this situation.

Instead, look back at the previous examples and notice that the intersections look “perpendicular.” If you draw a line that passes through the origin and the point of intersection, the line will be perpendicular to the tangent line.

Going back to our example with the point $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right),$ the line that passes through both $(0,0)$ and $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$ is the line \[y=\frac{1}{\sqrt{3}}x.\]

The line perpendicular to $y=\frac{1}{\sqrt{3}}x$ and that passes through $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$ is the line \[y=-\sqrt{3}\left(x-\frac{\sqrt{3}}{2}\right)+\frac{1}{2}.\]

This idea will work for any point on the unit circle. And in fact, the idea extends to any circle with any center.

Tangent Lines on Circles

Take $p$ to be a point on some circle $C$. Then $L$ is tangent to $C$ at $p$ if $L$ passes through $p$ and is perpendicular to the line that passes through the center of $C$ and $p$.

Practice doing this with the following example. It will be helpful to think of vectors.

In this first example, we work with a circle centered at $(0,0)$ but with radius not equal to $1$.

Example 1

Determine the equation of the line that is tangent to the circle $C$ at the point $(2,4)$, where $C$ is the circle that is given by the equation \[x^2+y^2=20.\]

The vector that moves points on the line $L$ that passes through $(0,0)$, the center of $C$, and the point $(2,4)$, is \[V=(2,4)-(0,0)=\langle 2,4\rangle.\] The vector perpendicular to $V$ is thus $V_{\perp}=\langle -4,2\rangle.$ So the $L_{\perp}$ that is perpendicular to $L$ and passes through $(2,4)$ is \[\begin{align*} y&=-\frac{2}{4}(x-2)+4\\ y&=-\frac{1}{2}(x-2)+4. \end{align*}\] Thus the line that is tangent to $C$ at the point $(2,4)$ is \[y=-\frac{1}{2}(x-2)+4.\]

In the next example, we work with a circle not centered at $(0,0)$.

Example 1

Determine the equation of the line that is tangent to the circle $C$ at the point $(4,-2)$, where $C$ is the circle that is given by the equation \[(x-3)^2+(y-1)^2=10.\]

The vector that moves points on the line $L$ that passes through $(3,1)$, the center of $C$, and the point $(4,-2)$, is \[V=(4,-2)-(3,1)=\langle 1,-3\rangle.\] The vector perpendicular to $V$ is thus $V_{\perp}=\langle 3,1\rangle.$ So the $L_{\perp}$ that is perpendicular to $L$ and passes through $(4,-2)$ is \[\begin{align*} y&=\frac{1}{3}(x-4)-2\\ \end{align*}\] Thus the line that is tangent to $C$ at the point $(2,4)$ is \[y=\frac{1}{3}(x-4)-2.\]

Now we turn our attention to an ellipse. An ellipse can be written as \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1,\] where $(h,k)$ is the center of the ellipse and $a$ and $b$ are constants.

An ellipse can be created by transforming the right circle:

\[\begin{align*} \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}&=1\\ \left(\frac{x-h}{a}\right)^2+\left(\frac{y-k}{b}\right)^2&=1\\ \left(\frac{1}{a}x-\frac{h}{a}\right)^2+\left(\frac{1}{b}y-\frac{k}{b}\right)^2&=1. \end{align*}\]

In this rewritten, the ellipse is just the circle centered at $\left(\frac{h}{a},\frac{k}{b}\right)$ with radius $1$ scaled in $x$-coordinate by $a$ and the $y$-coordinate by $b$.

Therefore, one can identify the line tangent to an ellipse if they find a line tangent to the right the circle and then scale it by the right amount to get the line tangent to the ellipse.

Tangent Lines on Ellipses

Take $p$ to be a point on some ellipse $E$. There are asymmetric scalings $X_a$ and $Y_b$ so that for some circle $C$, \[E=(X_a\circ Y_b)(C).\] Then $L$ is tangent to $E$ at $p$ if and only if the $(X_a\circ Y_b)^{-1}(L)$ is tangent to the circle $C$ at the point $(X_a\circ Y_b)^{-1}(p).$

Practice with the following example.

Example 3

Determine the equation of the line that is tangent to the ellipse $E$ at the point $(3,5)$, where $E$ is the ellipse that is given by the equation \[\frac{(x-1)^2}{4}+\frac{(y+1)^2}{36}=2.\]

The ellipse $E$ can be rewritten like this \[\left(\frac{1}{2}x-\frac{1}{2}\right)^2+\left(\frac{1}{6}y+\frac{1}{6}\right)^2=2.\] So $E$ is the asymmetric scaling of $2$ in the $x$-coordinate, $X_2$ and $6$ in the $y$-coordinate, $Y_2$ of the circle \[\left(x-\frac{1}{2}\right)^2+\left(y+\frac{1}{6}\right)^2=2.\] The point on $C$ that is the “equivalent” point $(3,5)$ on $E$ is then \[(Y_\frac{1}{6}\circ X_\frac{1}{2})(3,5)=\left(\frac{3}{2},\frac{5}{6}\right).\]

The vector that moves points on the line $L$ that passes through $\left(\frac{1}{2},-\frac{1}{6}\right)$, the center of $C$, and $\left(\frac{3}{2},\frac{5}{6}\right)$, the point that is on $C$, is \[V=\left(\frac{3}{2},\frac{5}{6}\right)-\left(\frac{1}{2},-\frac{1}{6}\right)=\langle 1,1\rangle.\] The vector perpendicular to $V$ is thus $V_{\perp}=\langle -1,1\rangle.$ So the $L_{\perp}$ that is perpendicular to $L$ and passes through $\left(\frac{3}{2},\frac{5}{6}\right)$ is \[\begin{align*} y&=-\left(x-\frac{3}{2}\right)+\frac{5}{6}\\ \end{align*}\] This is the line $L_C$ that is tangent to $C$ at $\left(\frac{3}{2},\frac{5}{6}\right)$.

To get the line that is tangent to $E$ at $(3,5)$, asymmetrically scale the line by $(X_2\circ Y_6).$

For any point $\left(x,-\left(x-\frac{3}{2}\right)+\frac{5}{6}\right)$ on $L_C$, we have \[\begin{align*} (X_2\circ Y_6)\left(x,-\left(x-\frac{3}{2}\right)+\frac{5}{6}\right)&=\left(2\cdot x,6\cdot \left(-\left(x-\frac{3}{2}\right)+\frac{5}{6}\right)\right)\\ &=\left(2x,-6\left(x-\frac{3}{2}\right)+5\right) &&\text{set $X=2x$ so that $\frac{X}{2}=x$}\\ &=\left(X,-6\left(\frac{X}{2}-\frac{3}{2}\right)+5\right)\\ &=\left(X,-\frac{6}{2}\left(X-3\right)+5\right)\\ &=\left(X,-3\left(X-3\right)+5\right). \end{align*}\]

Therefore, the line that is tangent to $E$ at $(3,5)$ is \[y=-3\left(x-3\right)+5.\]

Tangent Intersection of Quadratic Functions

Now we return to our earliest example: $f(x)=x^2-1$ and the point $(2,3).$ As we saw before, the line tangent to $L$ intersects $f$ at $(2,3)$ and that intersection only happens at that point.

This is observation generalizes for all quadratic polynomials. Here is how to phrase this statement in a way that makes identifying the line easier.

Tangent Lines on Quadratric Polynomials

Take $p$ to be a point on some quadratic polynomial $f$. The line $L$ is tangent to $f$ at the point $p$ if $f-L$ has only one root. In other words, the discriminant of $f-L$ must be zero for $L$ to be tangent to $f$ at $p.$

We will identify the formula for the right tangent line to $f(x)=x^2-1$ at the point $(2,3)$ in the next example. Before we do that, recall that the discriminant of a quadratic polynomial $ax^2+bx+c$ is $b^2-4ac.$

Example 4

Take $f(x)=x^2-1$. Find the line $L$ tangent to $f$ at the point $(2,3).$

The line $L=m(x-2)+3$ intersects $f$ at the point $(2,3)$. It will only be tangent if and only if $f-L$ only has one root. The function $f-L$ looks like this: \[\begin{align*}(f-L)(x)&=x^2-1-(m(x-2)+3)\\ &=x^2-1-(mx-2m+3)\\ &=x^2-1-mx+2m-3\\ &=x^2-mx+2m-4. \end{align*}\] The quadratic polynomial $f-L$ having only one root is equivalent to its discriminant equaling zero. The discriminant of $f-L$ is \[\begin{align*}(-m)^2-4(1)(2m-4)&=m^2-8m+16. \end{align*}\] The discriminant equals zero if and only if $m=4$. This is because it can be rewritten like this: \[\begin{align*}m^2-8m+16&=(m-4)^2. \end{align*}\]

Therefore, the right slope needed for $L$ is $m=4.$

So the line tangent to $f$ at $(2,3)$ is thus \[y=4(x-2)+3.\]

Here is another example.

Example 5

Take $f(x)=2x^2+3x-1$. Find the line $L$ tangent to $f$ at the point $(1,4).$

The line $L=m(x-1)+4$ intersects $f$ at the point $(1,4)$. It will only be tangent if and only if $f-L$ only has one root. The function $f-L$ looks like this: \[\begin{align*}(f-L)(x)&=2x^2+3x-1-(m(x-1)+4)\\ &=2x^2+3x-1-(mx-m+4)\\ &=2x^2+3x-1-mx+m-4\\ &=2x^2+(3-m)x+m-5. \end{align*}\] The quadratic polynomial $f-L$ having only one root is equivalent to its discriminant equaling zero. The discriminant of $f-L$ is \[\begin{align*}(3-m)^2-4(2)(m-5)&=m^2-14m+49. \end{align*}\] The discriminant equals zero if and only if $m=7$. This is because it can be rewritten like this: \[\begin{align*}m^2-14m+49&=(m-7)^2. \end{align*}\]

Therefore, the right slope needed for $L$ is $m=7.$

So the line tangent to $f$ at $(1,4)$ is thus \[y=7(x-1)+4.\]

Tangential Intersection of Polynomials Functions

The final thing we focus on is tangent line on polynomials. Unfortunately, the idea that a tangent line can only intersect the function, does not need to be true for polynomials.

Instead, the intersection of the tangent line and polynomial will look a certain way.

Degree N intersection

Take $a$ to be a real number and $f$ to be a polynomial. The intersection of a line $L$ and a polynomial $f$ at the point $(a,f(a))$ is a degree $n$ intersection if the difference $f-L$ satisfies these conditions:

There is a polynomial $E_a$ so that \[(f-L)(x)=(x-a)^nE_a(x)\]
The polynomial $E_a$ does not have a zero at $a$.

In other words, $f-L$ can be factored into two parts: a polynomial that has a root of order $n$ at $x=a$ and a polynomial that does not have a root at $x=a$.

This definition allows us to describe what to look for when determining what line will be tangent to a polynomial at a prescribed point $p$.

Tangent Line on Polynomials

For any polynomial $f$, the line $L$ is tangent to $f$ at the point $(a,f(a))$ if the intersection is a at least 2 degree order intersection. In other words, there is a polynomial $E_a$ so that \[f(x)=L(x)+(x-a)^2E_a(x).\]

The key to obtaining the tangent line is rewriting the polynomial $f$.

Replace $x$ with $a+(x-a)$ in $f$.
Expand $f(a+(x-a))$ as sums of products of factors that only involve $a$ and $(x-a)$
Rewrite this expansion to obtain, for some polynomial $E_a$ and some number $m$ the formula \[f(x)=f(a)+m(x-a)+(x-a)^2E_a(x).\]
The line $L(x)=m(x-a)+f(a)$ is the line tangent to $f$ at $f$ at $(a,f(a)).$

We will practice with a simple example first: a linear function.

Example 6

Take $f(x)=3x$. Determine an equation for the line $L$ that is tangent to $f$ at $(4,f(4)).$

First note that $f(4)=12.$ Expand $f(x)=f(4+(x-4))$ like this: \[\begin{align*} f(x)&=f(4+(x-4))\\ &=3(4+(x-4))\\ &=12+3(x-4). \end{align*}\] This is the tangent line to $f$ at $(4,12).$

In the above example, the slope of the line tangent to $f(x)=3x$ had the same slope. And in fact, it was the same line!

The above example generalizes. Any line tangent to a line will be itself.

Now we look at a simple quadratic.

Example 7

Take $f(x)=x^2$. Determine an equation for the line $L$ that is tangent to $f$ at $(3,f(3)).$

First note that $f(3)=9.$ Expand $f(x)=f(3+(x-3))$ like this: \[\begin{align*} f(x)&=f(3+(x-3))\\ &=\left(3+(x-3)\right)^2\\ &=9+6(x-3)+(x-3)^2. \end{align*}\] The line $L(x)=6(x-3)+9$ is the tangent line to $f$ at $(3,9).$

The pattern is not clear like in the previous example, so let us look at the next example that is more general.

Example 8

Take $f(x)=x^2$. Determine an equation for the line $L$ that is tangent to $f$ at $(a,f(a)).$

First note that $f(a)=a^2.$ Expand $f(x)=f(a+(x-a))$ like this: \[\begin{align*} f(x)&=f(a+(x-a))\\ &=\left(a+(x-a)\right)^2\\ &=a^2+2a(x-a)+(x-a)^2. \end{align*}\] The line $L(x)=2a(x-a)+a^2$ is the tangent line to $f$ at $(a,a^2).$

In the above example, the slope is given by $2a.$

Let us now revist Example 5 and see if the pattern is the same.

Example 9

Take $f(x)=2x^2+3x-1$. Determine an equation for the line $L$ that is tangent to $f$ at $(1,f(1)).$

First note that $f(1)=4.$ Expand $f(x)=f(1+(x-1))$ like this: \[\begin{align*} f(x)&=f(1+(x-1))\\ &=2(1+(x-1))^2+3(1+(x-1))-1\\ &=2(1+2(x-1)+(x-1)^2)+3(1+(x-1))-1\\ &=2+4(x-1)+2(x-1)^2+3+3(x-1)-1\\ &=4+7(x-1)+2(x-1)^2. \end{align*}\] The line $L(x)=7(x-1)+4$ is the tangent line to $f$ at $(1,4).$ This is the same answer as before.

This is the same answer we got in Example 5. Consistency is a good thing!

Now we try a cubic polynomial.

Example 10

Take $f(x)=3x^3+2x+4$. Determine an equation for the line $L$ that is tangent to $f$ at $(1,f(1)).$

First note that $f(1)=9.$ Expand $f(x)=f(1+(x-1))$ like this: \[\begin{align*} f(x)&=f(1+(x-1))\\ &=3(1+(x-1))^3+2(1+(x-1))+4\\ &=3(1+3(x-1)+3(x-1)^2+(x-1)^3)+2(1+(x-1))+4\\ &=3+9(x-1)+9(x-1)^2+3(x-1)^3+2+2(x-1)+4\\ &=9+11(x-1)+9(x-1)^2+3(x-1)^3\\ &=9+11(x-1)+(x-1)^2(9+3(x-1)). \end{align*}\] The line $L(x)=11(x-1)+9$ is the tangent line to $f$ at $(1,9).$

The calculations we have done are more or less similar. Therefore, it should be possible to generalize the process. To do this we use the following notation.

Slope of a Polynomial at a Point

For any polynomial $f,$ denote by $f'(a)$ to be the slope of the line that is tangent to $f$ at $(a,f(a)).$

For any two polynomials $f$ and $g$ and real number $a$, there are polynomials $F$ and $G$ so that

\[f(x)=f(a)+f'(a)(x-a)+F(x)(x-a)^2\] and \[g(x)=g(a)+g'(a)(x-a)+G(x)(x-a)^2.\]

Notice that for any real numbers $c_1$ and $c_2$, $c_1f(x)+c_2g(x)$ can be rewritten like this:

\[\begin{align*} c_1f(x)+c_2g(x)&=c_1f(a)+c_1f'(a)(x-a)+c_1F(x)(x-a)^2+c _2g(a)+c_2g'(a)(x-a)+c_2G(x)(x-a)^2\\ &=(c_1f(a)+c_2g(a))+(c_1f'(a)+c_2g'(a))(x-a)+(c_1F(x)+c_2G(x))(x-a)^2. \end{align*}\]

So therefore, the slope of line tangent to $c_1f(x)+c_2g(x)$ is $c_1f'(a)+c_2g'(a)$, which is the the sum of the slopes of $c_1f$ and $c_2g$ each scaled by the appropriate scaling factor.

Linearity of Slope of Tangent Line to Polynomial

For any polynomials $f$ and $g$ and any real number $c_1$ and $c_2$, the slope of the line tangent to $c_1f+c_2g$ at $a$ is \[(c_1f+c_2g)'(a)=c_1f'(a)+c_2g'(a).\] This is known as linearity.

Now we look at products:

For any two polynomials $f$ and $g$ and real number $a$, there are polynomials $F$ and $G$ so that

\[f(x)=f(a)+f'(a)(x-a)+F(x)(x-a)^2\] and \[g(x)=g(a)+g'(a)(x-a)+G(x)(x-a)^2.\]

Notice $f(x)g(x)$ can be rewritten like this:

\[\begin{align*} f(x)+g(x)&=\left(f(a)+f'(a)(x-a)+F(x)(x-a)^2\right)\cdot \left(g(a)+g'(a)(x-a)+G(x)(x-a)^2\right)\\ &=f(a)g(a)+(f'(a)g(a)+g'(a)f(a))(x-a)+B(x)(x-a)^2, \end{align*}\] where $B$ is some polynomial.

So therefore, the slope of line tangent to $f(x)g(x)$ is $f'(a)g(a)+f(a)g'(a)$.

Product Rule for Slope of Tangent Line to Polynomial

For any polynomials $f$ and $g$, the slope of the line tangent to $fg$ at $a$ is \[(fg)'(a)=f'(a)g(a)+f(a)g'(a).\] This is known as the product rule.

By combing the previous results, we have the following theorem:

Power Rule

For any natural number $n$ greater than $1$, we have that

${\rm{pow}}'_n(a)=na^{n-1}$
${\rm{pow}}'_1(a)=1$
If $f(x)=1$, then $f'(a)=0.$

Practice using the Power Rule, Product Rule, and Linearity to find the tangent line of the following polynomial.

Example 10

Take $f(x)=-4x^4+8x^3+x^2+5x+1$. Determine an equation for the line $L$ that is tangent to $f$ at $(1,f(1)).$

First note that $f(1)=11.$ The function $f$ is the sum of power functions:

\[f=-4{\rm{pow}}_4+8{\rm{pow}}_3+{\rm{pow}}_2+5{\rm{pow}}_1+1\] so the slope of its tangent line can be computed like this:

\[\begin{align*} f'(a)&=-4({\rm{pow}}_4)'(a)+8({\rm{pow}}_3)'(a)+({\rm{pow}}_2)'(a)+5({\rm{pow}}_1)'(a)+(1)'\\ &=-4\cdot 4a^3+8\cdot 3a^2+2a+5+0\\ &=-16a^3+24a^2+2a+5. \end{align*}\]

Evaluate this at $a=1$ to get that

\[f'(1)=15.\]

Therefore the line tangent to $f$ at $(1,11)$ is

\[y=15(x-1)+11.\]

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