Chapter 4.4 Symmetry of Tangential Intersections
In this section, we revisit tangential intersections by indentifying what kind of symmetry these special kind of intersections have. We will also discuss tangential intersection of ellipses.
Translating Intersection
Recall that a line is tangent to a quadratic function or a circle at a point \(p\) if and only if the point \(p\) is the unique point of intersection of the line with the quadratic function or the circle.
First we focus on quadratic functions. It turns out that translating the plane doe not change intersection. Therefore, tangency has translation symmetry.
We will show that this is true for quadratic functions.
Example 1
Take \(L_a\) to be the line tangent to \(\rm{pow}_2\) at the point \((a,a^2)\). Show that for any non-zero real number \(A\), and any real numbers \(h\) and \(k\), the line \(\langle h,k\rangle +Y_AL_a\) is tangent to \(\langle h,k\rangle +Y_A\rm{pow}_2\) at the point \((a+h,A\rm{pow}_2(a)+k).\)
The line \(L_a\) is tangent to \(\rm{pow}_2\) and intersects it at only the point \((a,a^2).\) Because asymmetric scalings in the \(y\)-coordinate, \(Y_A\), preserves intersections, the line \(Y_A L_a\) intersects \(Y_A\rm{pow}_2\) at exactly one point. Because translation by the vector \(\langle h,k\rangle\) also preserves interse tions, the line \(\langle h,k\rangle+Y_AL_a\) intersects \(\langle h,k\rangle+Y_A{\rm{pow}}_2\) at exactly one point. We have that \[\begin{align*} \langle h,k\rangle+Y_A(a,a^2)&=\langle h,k\rangle+(a,Aa^2)\\ (a+h,Aa^2+k). \end{align*}\] So the line \(\langle h,k\rangle+Y_AL_a\) intersects \(\langle h,k\rangle+Y_A\rm{pow}_2\) at the point \((a+h,Aa^2+k)\). Now we derive the equation of the new tangent line. The line \(L_a\) looks like this: \[y=2a(x-a)+a^2.\] So the line \(L=\langle h,k\rangle+Y_AL_a\) looks like this: We have that \[\begin{align*} \langle h,k\rangle+Y_AL_a(x)&=A(2a(x-h-a)+a^2)+k\\ &=2aA(x-(a+h))^2+Aa^2+k. \end{align*}\] The function \(f=\langle h,k\rangle+Y_A\rm{pow}_2\) is given by the quation \(f(x)=A(x-h)^2+k,\) so \(L\) is tangent to \(f\) at \((a+h,Aa^2+k)\) and the slope of the line is \(2aA\).
In this next example, use the given information about a quadratic polynomial to identify the equation of the line tangent it.
Example 2
Take \(L\) to be the line with slope \(3\) that intersects \((1,5)\) and is tangent to the quadratic polynomial \(f\) at \((1,5),\) Identify a point \(p\) and a line \(R\) so that \(R\) is tangent to the quadratic polynomial \(g\) at \(p\), where \[g(x)=4f(x-1)+9.\]
The function \(g\) can be written like this: \[g=\langle 1,9\rangle +Y_4 f.\] So the line \(R\) given by \[R=\langle 1,9\rangle +Y_4L\] is tangent to \(g\) at the point \[\langle 1,9\rangle +Y_4(1,5)=(2,29).\] The slope of the line \(R\) is \[4\cdot 3=12.\] So the line \(R\) given by \[y=12(x-2)+29\] and it is tangent to \(g\) at the point \((2,29).\)
What we discussed with quadratic polynomials also applies to polynomials in general. That is Translation preserves tangency.
Example 3
Take \(f\) to be the polynomial given by \[f(x)=x^3-2x-3.\]
- Identify an equation for the line tangent to \(f\) at the point \((2,1).\)
- Use the invariance of tangency under translation to identify an equation for the line tangent to \(g\) at the point \((3,5)\), where \(g\) is the polynomial that is given by \[g(x)=f(x-1)+4.\]
- Write out \(g\) explicitly and determine the line directly from this formula for \(g.\)
Expand \(f\) by writing \(x\) as \(2+(x-2)\): \[\begin{align*} f(x)&=f(2+(x-2))\\ &=(2+(x-2))^3-2(2+(x-2))-3\\ &=(8+12(x-2)+6(x-2)^2+(x-2)^3)-2(2+(x-2))-3\\ &=(8-4-3)+(12-2)(x-2)+(6+(x-2))(x-2)^2. \end{align*}\] The line tangent to \(f\) at \((2,1)\) is \[L(x)=(8-4-3)+(12-2)(x-2)=1+10(x-2)\] which is equivalent to \[L(x)=10x-19.\]
The function \(g\) can be rewritten like this \[g=\langle 1,4\rangle +f,\] so \[R=\langle 1,4\rangle +L\] is tangent to \(g\) at \[\langle 1,4\rangle +(2,1)=(3,5).\] The formula is \[\begin{align*} R(x)&=L(x-1)+4\\ &=10(x-1)-19+4\\ &=10x-25. \end{align*}\]
The formula for \(g\) is \[\begin{align*} g(x)&=f(x-1)+4\\ &=(x-1)^3-2(x-1)+1. \end{align*}\] Expand \(g\) around \(3\) to get \[\begin{align*} g(x)&=(2+(x-3))^3-2(2+(x-3))+1\\ &=(8+12(x-3)+6(x-3)^2+(x-3)^2)-2(2+(x-3))+1\\ &=8-4+1+(12-2)(x-3)+(6+(x-3))(x-3)^2\\ &=5+10(x-3)+(6+(x-3))(x-3)^2\\ &=10x-25+E(x)(x-3)^2. \end{align*}\] The tangent line is \(R(x)=10x-25\) which is the same answer as we got in b.
Scaling Intersections
We saw that translation preserves tangency. What about asymmetric scalings? Do they preserve tangency?
It turns out that they do!
Study this with the next example.
Example 4
Take \(f\) to be the polynomial given by \[f(x)=x^3-2x-3.\]
- Identify an equation for the line tangent to \(f\) at the point \((2,1).\)
- Use the invariance of tangency under \(y\)-axis and \(x\)-axis scaling to identify an equation for the line tangent to \(g\) at the point \((1,5)\), where \(g\) is the polynomial that is given by \[g(x)=5f(2x).\]
- Write out \(g\) explicitly and determine the line directly from this formula for \(g.\)
We did this already by expanding around \(2\). The answer is \[L(x)=10x-19.\]
The function \(g\) can be rewritten like this \[g=S_5\circ f\circ S_2.\] So \[R=S_5\circ L\circ S_2\] is tangent to \(g\) at \[(\frac{1}{2}\cdot 2,1\cdot 5)=(1,5).\] The formula is \[\begin{align*} R(x)&=(S_5\circ L\circ S_2)(x)\\ &=5\cdot L(2x)\\ &=5(10(2x)-19)\\ &=5(20x-19)\\ &=100x-95. \end{align*}\]
The formula for \(g\) is \[\begin{align*} g(x)&=5f(2x)\\ &=5((2x)^3-2(2x)-3)\\ &=40x^3-20x-15. \end{align*}\] Expand \(g\) around \(1\) to get \[\begin{align*} g(x)&=40(1+(x-1))^3-20(1+(x-1))-15\\ &=40(1+3(x-1)+3(x-1)^2+(x-1)^2)-20(1+(x-1))-15\\ &=40-20-15+(120-20)(x-1)+(3+(x-1)^2)(x-1)^2\\ &=5+100(x-1)+(3+(x-3))(x-1)^2\\ &=100x-95+E(x)(x-1)^2. \end{align*}\] The tangent line is \(R(x)=100x-95\) which is the same answer as we got in b.
An example of a set that is assymmetrically scaled is an ellipse.
First, recall how tangent line works for a circle.
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 \(p\).
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 5
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 \[y=-3\left(x-3\right)+5.\]
Here is another example.
Example 6
Take \(E\) is the ellipse that is given by the equation \[\frac{(x-1)^2}{9}+\frac{(y-3)^2}{4}=1.\] Identify an equation for the line tangent to \(E\) at the point \((\frac{14}{5},\frac{23}{5}).\)
The ellipse can be transformed into the unit circle by translating it by the vector \(\langle -1, -3\rangle\) and then assymmetrically scaling it by \(X_\frac{1}{3}\circ Y_\frac{1}{2}\): \[\begin{align*} \langle -1,-3\rangle + E&=\left\{(x-1,y-3)\colon \frac{(x-1)^2}{9}+\frac{(y-3)^2}{4}=1\right\}\\ &=\left\{(X,Y)\colon \frac{X^2}{9}+\frac{Y^2}{4}=1 \right\}&&\text{where }X=x-1\text{ and }Y=y-3.\\ \end{align*}\] Now assymmetrically scale to get \[\begin{align*} X_\frac{1}{3}\circ Y_\frac{1}{2}(\langle -1,-3\rangle+E)&=\left\{\left(\frac{x}{3},\frac{y}{2}\right)\colon \frac{x^2}{9}+\frac{x^2}{4}=1 \right\}\\ &=\left\{(X,Y)\colon \frac{(3X)^2}{9}+\frac{(2Y)^2}{4}=1 \right\}&&\text{where }X=\frac{x}{3}\text{ and }Y=\frac{y}{2}.\\ &=\left\{(X,Y) \colon X^2+Y^2=1 \right\}. \end{align*}\] So we get the unit circle \(\mathcal{C}\). Conversely, if we assymetrically scale by \(X_3Y_2\) the unit circle and then translate it by \(\langle 1,3\rangle\), we get \(E\). That is \[E=\langle 1,3\rangle +X_3 Y_2 \mathcal{C}.\] So, we can first find the tangent line to \(\mathcal{C}\) at \[X_\frac{1}{3}Y_\frac{1}{2}\left(\langle -1,-3\rangle +\left(\frac{14}{5},\frac{23}{5}\right)\right)=X_\frac{1}{3}Y_\frac{1}{2}\left(\frac{9}{5},\frac{8}{5}\right)=\left(\frac{3}{5}.\frac{4}{5}\right).\] The line tangent to \(\mathcal{C}\) at \(\left(\frac{3}{5}.\frac{4}{5}\right)\) is the line \[L(x)=-\frac{3}{4}\left(x-\frac{3}{5}\right)+\frac{4}{5}=-\frac{3}{4}x+\frac{25}{20}.\] So the line tangent to \(E\) at \((\frac{14}{5},\frac{23}{5})\) is given by \[\begin{align*} R(x)&=\left[(\langle 1,3\rangle +X_3Y_2)L\right](x)\\ &=2L\left(\frac{1}{3}(x-1)\right)+3\\ &=2\left(-\frac{3}{4}\cdot \frac{1}{3}(x-1)+\frac{25}{20}\right)+3\\ &=2\left(-\frac{1}{4}x+\frac{30}{20}\right)+3\\ &=-\frac{1}{2}x+3+3\\ &=-\frac{1}{2}x+6. \end{align*}\]
An Algebraic Inverse Function Theorem
Now we discuss tangency for inverse functions.
Recall that for any line \(L\) in the plane, reflection across \(L\) preserves distance and so should preserve the properties of intersections.
That means that for any polynomial or rational function \(f\), if \(L\) is a line that is tangent to \(f\) at \((a,f(a)),\) then the reflection of \(L\) across \(\rm{pow}_1\) should be tangent to \(f^{-1}\) at \((f(a),a)\).
Specifically, if \(L\) is tangent to \(f\) at \((a,f(a))\), then it is given by the equation
\[y=f'(a)(x-a)+f(a).\] The reflection of \(L\) across the line \(\rm{pow}_1\) is given by \[y=\frac{1}{f'(a)}(x-f(a))+a.\]
Therefore,
\[(f^{-1})'(f(a))=\frac{1}{f'(a)}.\]
Use this information to answer the following problem.
Example 7
Take \(f\) to be the restriction of \(\rm{pow}_2\) to \([0,\infty).\) The function \(f\) is invertible and \(f^{-1}\) is the square root function.
- Determine the line \(L\) that is tangent to \(f\) at \((2,4)\).
- Reflect \(L\) and \(f\) across \(\rm{pow}_1\) to obtain \(L^{-1}\), \(f^{-1}\), and their intersection at \((4,2)\).
- Determine \((f^{-1})'(4).\)
We have that \({\rm{pow}_2}'(2)=2\cdot 2=4.\) So, \[L(x)=4(x-2)+4=4x-4.\]
The inverse of \(f\) is \[f^{-1}(x)=\sqrt{x}\] and \[L^{-1}(x)=\frac{1}{4}(x-4)+2.\]
Here \(f(2)=4\), so this will equal the reciprocal of the derivative evaluated at \(2\): \[(f^{-1})'(4)=\frac{1}{4}.\]
In this next example, we will evaluate the derivative of the inverse without finding a formula for the inverse!
Example 8
Evaluate \((f^{-1})^\prime(4)\), where \(f\) is the function given by \[f(x)=x^3+2x+1.\] Note that \(f(1)\) is equal to \(4.\)
Because \(f(1)=4\). We are trying to determine \[(f^{-1})^\prime(4)=(f^{-1})^\prime(f(1)).\] So, the answer will be the reciprocal of the derivative of \(f\) evaluated at \(1\). The derivative of \(f\) is \[f^\prime (x)=3x^2+2,\quad \text{so}\quad f^\prime(1)=5.\] Therefore, \[(f^{-1})^\prime(4)=\frac{1}{5}.\]
We can use what we learned to find the derivative of the inverse of \({\rm pow}_n\).
Assume that \(n\) is an odd natural number. Then \({\rm pow}_n\) is invertible and its inverse is \({\rm pow}_\frac{1}{n}\). Moreover, for any \((a,a^n)\) in \({\rm pow}_n\), we have that
\[{\rm pow}_n^\prime (a)=na^{n-1}.\]
And so, for any non-zero \(a\),
\[{\rm pow}_{\frac{1}{n}}^\prime (a^n)=\frac{1}{{\rm pow}_n^\prime (a)}=\frac{1}{na^{n-1}},\] and so
\[{\rm pow}_{\frac{1}{n}}^\prime(a)=\frac{1}{n\left(a^\frac{1}{n}\right)^{n-1}}=\frac{1}{n} a^{\frac{1}{n}-1}.\]
Therefore, for any odd natural number \(n\) and non-zero real number \(a\),
\[{\rm pow}_{\frac{1}{n}}^\prime(a)=\frac{1}{n} a^{\frac{1}{n}-1}.\]
This also is true if \(n\) is an even number, except \(a\) must be positive so that \({\rm{pow}}_n\) is invertible.
Example 9
Identify an equation for the line that is tangent to \(\rm{pow}_\frac{1}{3}\) at the point \((27,3).\)
We have that \[{\rm pow}_{\frac{1}{3}}^\prime(27)=\frac{1}{3} (27)^{\frac{1}{3}-1}=\frac{1}{3}(27)^{-\frac{2}{3}}=\frac{1}{27}.\]
So the equation of the tangent line is \[y=\frac{1}{27}(x-27)+3.\]