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There’s a little chance that you clicked this because of the title, which is somewhat subtle, given that we’re not marketing anything. So, if this turned out as a clickbait, we apologize straightway; but please wait, read on.

In our previous post, we talked about linear-quadratic systems. However, as that post was getting longer than we wanted, and we were gradually going off on a tangent, we decided to separate this content. Ironically, we went off on a tangent not long after we started talking about \cdots tangents. Can anything be closer to an irony?

Tangent conditions for conic sections

The result below has been proved for the circle x^2+y^2=r^2 and the parabola y=x^2; we now consider ellipses and hyperbolas.

Proposition (ellipse)

If the linear-quadratic system

(1)   \begin{equation*} y=mx+k \end{equation*}

(2)   \begin{equation*} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \end{equation*}

has only one point of intersection, then:

  1. k^2=a^2m^2+b^2
  2. the point of tangency has coordinates (\frac{-kma^2}{a^2m^2+b^2},\frac{kb^2}{a^2m^2+b^2}).

To prove the first statement, we utilize the fact that there’s just one point of intersection, which in turn means that the discriminant of the resulting quadratic is zero. Substitute mx+k for y in equation (2):

    \begin{equation*} \begin{split} \frac{x^2}{a^2}+\frac{(mx+k)^2}{b^2}&=1\\ \frac{x^2}{a^2}+\frac{m^2x^2+2kmx+k^2}{b^2}&=1\\ b^2x^2+a^2(m^2x^2+2kmx+k^2)=a^2b^2&\\ b^2x^2+a^2m^2x^2+2a^2kmx+a^2k^2=a^2b^2&\\ (a^2m^2+b^2)x^2+(2a^2km)x+a^2k^2-a^2b^2=0&\\ \vdots&\\ (2a^2km)^2-4(a^2m^2+b^2)(a^2k^2-a^2b^2)&=0\\ 4a^4k^2m^2-4(a^4k^2m^2-a^4b^2m^2+a^2b^2k^2-a^2b^4)&=0\\ 4a^4k^2m^2-4a^4k^2m^2+4a^4b^2m^2-4a^2b^2k^2+4a^2b^4&=0\\ 4a^4b^2m^2-4a^2b^2k^2+4a^2b^4&=0\\ a^4b^2m^2-a^2b^2k^2+a^2b^4&=0\\ a^4b^2m^2+a^2b^4=a^2b^2k^2&\\ a^2b^2(a^2m^2+b^2)=a^2b^2k^2&\\ a^2m^2+b^2=k^2, \end{split} \end{equation}

as required. This proves the first statement. To prove the second statement, we note that the only root of the quadratic equation

    \[(a^2m^2+b^2)x^2+(2a^2km)x+a^2k^2-a^2b^2=0\]

is at its vertex, namely x=-\frac{2a^2km}{2(a^2m^2+b^2)}=-\frac{a^2km}{a^2m^2+b^2}, which is the x-coordinate of the tangent point. Next, we use equation (1) to get the y-coordinate of the tangent point:

    \begin{equation*} \begin{split} y&=mx+k\\ &=m\Big(-\frac{a^2km}{a^2m^2+b^2}\Big)+k\\ &=m\Big(-\frac{a^2km}{a^2m^2+b^2}\Big)+k\Big(\frac{a^2m^2+b^2}{a^2m^2+b^2}\Big)\\ &=\frac{-a^2km^2}{a^2m^2+b^2}+\frac{a^2km^2+kb^2}{a^2m^2+b^2}\\ &=\frac{kb^2}{a^2m^2+b^2} \end{split} \end{equation}

This verifies that the point of tangency is at (\frac{-kma^2}{a^2m^2+b^2},\frac{kb^2}{a^2m^2+b^2}).

For a circle, a=b; so the above result simplifies to what we got for the case of circles.

Proposition (unit hyperbola)

If the linear-quadratic system

(3)   \begin{equation*} y=mx+k \end{equation*}

(4)   \begin{equation*} x^2-y^2=1 \end{equation*}

has only one point of intersection, then:

  1. m^2-k^2=1;
  2. the point of tangency has coordinates \Big(\frac{mk}{1-m^2},\frac{k}{1-m^2}\Big).

The proof is similar to the case of an ellipse, so is omitted.

Obtaining three free tangents from one tangent

Now comes our main point, which motivated the title of this post. It has to do with circles, ellipses, and hyperbolas being generally generous, when it comes to tangents.

Proposition

For the linear-quadratic system

(5)   \begin{equation*} y=mx+b,\quad\quad x^2+y^2=r^2 \end{equation*}

the following statements are equivalent:

  1. the line y=mx+b is a tangent;
  2. the line y=mx-b is a tangent;
  3. the line y=-mx+b is a tangent;
  4. the line y=-mx-b is a tangent.

In other words, one tangent gives rise to three other tangents.

To prove the above equivalence, we first substitute mx+b for y in the second equation, just like we did previously and obtain the quadratic equation

    \[(1+m^2)x^2+(2bm)x+(b^2-r^2)=0,\]

whose discriminant is (2bm)^2-4(1+m^2)(b^2-r^2). Now, if the first claim is true, then

    \begin{equation*} \begin{split} 0&=(2bm)^2-4(1+m^2)(b^2-r^2)\\ &=\Big(2(-b)m\Big)^2-4(1+m^2)\Big((-b)^2-r^2\Big)\\ &=\Big(2b(-m)\Big)^2-4(1+(-m)^2)\Big(b^2-r^2\Big)\\ &=\Big(2(-b)(-m)\Big)^2-4(1+(-m)^2)\Big((-b)^2-r^2\Big), \end{split} \end{equation}

and so (1)\implies (2)\implies (3)\implies (4)\implies (1).

Proposition

For the linear-quadratic system

(6)   \begin{equation*} y=mx+k,\quad\quad \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \end{equation*}

the following statements are equivalent:

  1. the line y=mx+k is a tangent;
  2. the line y=mx-k is a tangent;
  3. the line y=-mx+k is a tangent;
  4. the line y=-mx-k is a tangent.

The proof is just the same as in the case of a circle, so we skip it.

Proposition

For the linear-quadratic system

(7)   \begin{equation*} y=mx+k,\quad\quad x^2-y^2=1 \end{equation*}

the following statements are equivalent:

  1. the line y=mx+k is a tangent;
  2. the line y=mx-k is a tangent;
  3. the line y=-mx+k is a tangent;
  4. the line y=-mx-k is a tangent.

Again we omit the proof, since it is similar to the case of a circle.

Graphical illustration (circle case)

Rendered by QuickLaTeX.com


Exercises for the reader

  1. If the linear-quadratic system

        \[y=mx+k,\quad \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\]

    has only one point of intersection, prove that:

    • k^2+a^2(1-m^2)=0;
    • the point of tangency is \Big(\frac{a^2km}{b^2-a^2m^2},\frac{b^2k}{b^2-a^2m^2}\Big).
  2. Prove that a line with a slope m “sandwiched” between -1 and 1 can never be a tangent to a hyperbola. In other words, if -1< m < 1, then it is impossible for the line y=mx+k to be a tangent to the hyperbola \frac{x^2}{a^2}-\frac{y^2}{b^2}=1.
    [What about when m=\pm 1? You may use the first part of the preceding exercise for the case -1< m < 1.]