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## Buy one, get three free!

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 tangents. Can anything be closer to an irony?

## Tangent conditions for conic sections

The result below has been proved for the circle and the parabola ; we now consider ellipses and hyperbolas.

#### Proposition (ellipse)

If the linear-quadratic system

(1)

(2)

has only one point of intersection, then:

1. the point of tangency has coordinates .

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 for in equation (2):

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

is at its vertex, namely , which is the -coordinate of the tangent point. Next, we use equation (1) to get the -coordinate of the tangent point:

This verifies that the point of tangency is at .

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

#### Proposition (unit hyperbola)

If the linear-quadratic system

(3)

(4)

has only one point of intersection, then:

1. ;
2. the point of tangency has coordinates .

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)

the following statements are equivalent:

1. the line is a tangent;
2. the line is a tangent;
3. the line is a tangent;
4. the line is a tangent.

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

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

whose discriminant is . Now, if the first claim is true, then

and so .

#### Proposition

For the linear-quadratic system

(6)

the following statements are equivalent:

1. the line is a tangent;
2. the line is a tangent;
3. the line is a tangent;
4. the line 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)

the following statements are equivalent:

1. the line is a tangent;
2. the line is a tangent;
3. the line is a tangent;
4. the line is a tangent.

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

## Exercises for the reader

1. If the linear-quadratic system

has only one point of intersection, prove that:

• ;
• the point of tangency is .
2. Prove that a line with a slope “sandwiched” between and can never be a tangent to a hyperbola. In other words, if , then it is impossible for the line to be a tangent to the hyperbola .
[What about when ? You may use the first part of the preceding exercise for the case .]

As the name implies, a linear-quadratic system is made up of two equations in which one is linear and the other is quadratic. In simplest form, such a system may be given by

Questions related to linear-quadratic systems may not be presented in the above form; in that case, a preliminary re-arrangement will be necessary. However, if we suppose that we’re working with the system above, then an algebraic solution proceeds by substitution:

The resulting one-variable equation being quadratic indicates that one should be able to solve quadratic equations before attempting problems on linear-quadratic systems; otherwise, it will be a challenging undertaking.

## Possible scenarios

There are three possibilities that arise when solving linear-quadratic systems:

• two solutions: the line can intersect the quadratic in two distinct points (if this is the case, the line is called a secant);

• one solution: the line can intersect the quadratic in just one point (if this is the case, the line is called a tangent to the quadratic);

• no solution: the line doesn’t intersect the quadratic.

#### Example 1

Solve the linear-quadratic system

(1)

(2)

Substitute the right side of equation (2) into the left side of equation (1): , and then continue the solution:
\begin{split}
x^2-2x+3&=x+1\\
x^2-2x-x+3-1&=0\\
x^2-3x+2&=0\\
(x-1)(x-2)&=0\\
x&=1~\textrm{or}~x=2\\
y&=x+1\\
\end{split}
Therefore, the line intersects the quadratic at the points and . The line is a secant in this case.

#### Example 2

Solve the linear-quadratic system

(3)

(4)

As usual, we replace the left side of equation (3) with the right side of equation (4), and then solve the resulting quadratic equation:
\begin{split}
3x^2-4x+2&=2x-1\\
3x^2-4x-2x+2+1&=0\\
3x^2-6x+3&=0\\
3(x^2-2x+1)&=0\\
3(x-1)^2&=0\\
&\vdots\vdots\vdots\vdots\vdots\\
y&=2x-1\\
y&=2\times 1-1\\
y&=1
\end{split}
So, . The line is a tangent in this case. You can verify this by graphing both equations.

#### Example 3

Solve the linear-quadratic system

(5)

(6)

In this case, it is more convenient to isolate from equation (6), and then substitute in equation (5). From equation (6) we have:

(7)

Substitute the right side of equation (7) for in equation (5), simplify, and solve:
\begin{split}
3x+2\boxed{\frac{x^2}{3}}+3&=0\\
3x+2\Big(\frac{x^2}{3}\Big)+3&=0\\
9x+2x^2+9&=0\\
2x^2+9x+9&=0\\
2x^2+6x+3x+9&=0\\
2x(x+3)+3(x+3)&=0\\
(x+3)(2x+3)&=0\\
x=-3~\textrm{or}~x=-\frac{3}{2}&\\
&\vdots\vdots\vdots\vdots\vdots\\
y&=\frac{x^2}{3}\\
y&=\frac{(-3)^2}{3}=\frac{9}{3}=3\\
\textrm{or}~y&=\frac{(-3/2)^2}{3}=\frac{9/4}{3}=\frac{3}{4}
\end{split}
There are two points of intersection in this case, namely and . You can also verify this by graphing.

#### Example 4

Solve the linear-quadratic system

(8)

(9)

We have
\begin{split}
x^2&=-x-1\\
x^2+x+1&=0\\
x&=\frac{-1\pm\sqrt{1^2-4\times 1\times 1}}{2\times 1}\\
x&=\frac{-1\pm\sqrt{-3}}{2}\\
\end{split}
Since the value obtained for is not real, the linear-quadratic system has no real solution.

## Condition for tangency in a linear-quadratic system

A special case occurs when the line is a tangent to the quadratic , and it comes with a special requirement — the discriminant of the quadratic that results from the substitution has to be zero.

#### Example 5

The quadratic is to intersect a line with slope at just one point. Determine the conditions on the -intercept of the line.

Suppose that the -intercept of the line is . Since its slope is , its equation will be . Together with the given quadratic, we have the following linear-quadratic system:

(10)

(11)

Then, we proceed as usual:
\begin{split}
x^2-3x+1&=2x+k\\
x^2-5x+1-k&=0\\
&\vdots\vdots\vdots\vdots\vdots\\
\textrm{The discriminant must be zero for a tangent, so}&\\
(-5)^2-4(1)(1-k)&=0\\
25-4+4k&=0\\
21+4k&=0\\
k&=-\frac{21}{4}
\end{split}
The -intercept of the line must be .

## Tangents to the parent parabola: Buy one, get one free!

The tangent requirement between and is extremely simple.

#### Example 6

If is a tangent to , prove that:

1. ;
2. the point of tangency is ;
3. there’s a “free” associated tangent , whose point of tangency is at .
[In fact, is a tangent to if, and only if, is a tangent to .]

Set the right sides of the given equations equal to each other and continue the simplification:
\begin{split}
x^2&=mx+b\\
x^2-mx-b&=0\\
\vdots\\
m^2+4b&=0.
\end{split}
This proves the first statement. For the second statement, we will first calculate the -coordinate of the point of tangency by solving the quadratic equation . However, bearing in mind that the discriminant is zero, the solution will just be . Next, we obtain the -coordinate of the tangent point using :
\begin{split}
y&=mx+b\\
&=m\Big(\frac{m}{2}\Big)+b\\
&=\frac{m^2}{2}+b\\
&=\frac{m^2}{2}+\frac{2b}{2}\\
&=\frac{m^2+2b}{2}\\
&=-b
\end{split}
So we get that the point of tangency is , as desired. For the third statement, we “re-solve” the linear-quadratic system

and obtain the quadratic . But then, its discriminant is

which is the same discriminant as in the case of the earlier tangent . So is also a tangent. The -coordinate of the tangent point is obtained by solving , and it is . The -coordinate is then obtained by using in (or by symmetry of the parabola); it is . So the point of tangency between the second tangent and the parabola is .

## Tangents to a circle

Recall that the equation of a circle with center at the origin and radius is given by . Because of the symmetric nature of this quadratic equation, the tangent requirement for a circle also simplifies considerably.

#### Example 7

Prove that if the linear-quadratic system

(12)

(13)

has only one point of intersection, then:

1. ;
2. the point of tangency is .

Note that both statements can be proved by analytic geometry using the fact that radius is perpendicular to tangent at the point of contact, but right now we’re treating it as a linear-quadratic system problem. Let’s prove the first statement. Substitute for in equation (13) and obtain
\begin{split}
x^2+(mx+b)^2&=r^2\\
x^2+m^2x^2+(2bm)x+b^2&=r^2\\
(1+m^2)x^2+(2bm)x+(b^2-r^2)&=0\\
\end{split}
Now comes the main point. The discriminant of the quadratic equation above has to be zero. With this in mind, we have:
\begin{split}
(2bm)^2-4(1+m^2)(b^2-r^2)&=0\\
4b^2m^2-4(b^2-r^2+b^2m^2-m^2r^2)&=0\\
4b^2m^2-4b^2+4r^2-4b^2m^2+4m^2r^2&=0\\
-4b^2+4r^2+4m^2r^2&=0\\
-b^2+r^2+m^2r^2&=0\\
r^2+m^2r^2&=b^2\\
r^2(1+m^2)&=b^2,
\end{split}
as required. Next, we prove the second statement. To do this we just have to solve the quadratic equation under the assumption that its discriminant is zero. Let’s use the quadratic formula (although this is unnecessary since the discriminant is zero):
\begin{split}
x&=\frac{-2bm\pm\sqrt{(2bm)^2-4(1+m^2)(b^2-r^2)}}{2(1+m^2)}\\
&=\frac{-2bm}{2(1+m^2)}\quad\textrm{the square root above is zero}\\
&=\frac{-bm}{1+m^2}
\end{split}
This gives the -coordinate of the tangent point. To get the -coordinate, we use equation (12):
\begin{split}
y&=mx+b\\
&=m\Big(\frac{-bm}{1+m^2}\Big)+b\\
&=\frac{-bm^2}{1+m^2}+b\\
&=\frac{-bm^2}{1+m^2}+b\Big(\frac{1+m^2}{1+m^2}\Big)\\
&=\frac{-bm^2+b+bm^2}{1+m^2}\\
&=\frac{b}{1+m^2}
\end{split}
So, the tangent point has coordinates . This proves the second statement.

#### Example 8

Prove that if a line with slope is a tangent to the circle , then the point of tangency is precisely the midpoint of the and intercepts of the line.

From the preceding example, if a line is tangent to , we calculated that the point of tangency is . Now, when , this simplifies to . But then the intercept of the line is at , while the -intercept is at . By the midpoint formula, the midpoint of these intercepts is at

which is the same as the point of tangency.

## A simple word problem

As in the case of linear systems, there are certain word problems that lead to linear-quadratic systems. We provide an example below.

#### Example 9

Find two numbers whose sum and product are and , respectively.

Let the numbers be and . Since the sum of the numbers is , we have that ; since their product is , we have . Together, we have the linear-quadratic system:

(14)

(15)

Notice that the second equation is indeed quadratic, due to the term having a degree of : and each contribute an exponent of , making the overall exponent to be . From equation (14), we have . We then substitute for in equation (15) and obtain . Expanding and simplifying:
\begin{split}
27x-x^2&=182\\
27x-x^2-182&=0\\
-x^2+27x-182&=0\\
\textrm{or,}~x^2-27x-182&=0\\
(x-13)(x-14)&=0\\
\end{split}
We get or . Then, since , we equally get when , or when . Therefore the two numbers are and . Check that and .

## Trigonometric application of linear-quadratic systems

It shouldn’t come as a surprise that an important case of a linear-quadratic system occurs in trigonometry, where the unit circle is used to derive trigonometric ratios for some special angles.

#### Example 10

Solve the linear-quadratic system

(16)

(17)

Putting in equation (17) we have:
\begin{split}
x^2+x^2&=1\\
2x^2&=1\\
x^2&=\frac{1}{2}\\
x&=\pm\sqrt{\frac{1}{2}}\\
\textrm{or,}~x&=\pm\frac{\sqrt{2}}{2}\\
\vdots\vdots&\vdots\vdots\vdots\\
y&=x\\
\textrm{so,}~y&=\pm\frac{\sqrt{2}}{2}
\end{split}
This is how the trigonometric ratios for angle are derived using the unit circle. For angle or , see the exercises at the end.

## Exercises for the reader

1. If the linear-quadratic system

(18)

(19)

has only one point of intersection, prove that:

• ;
• the point of tangency is .

Observe that if , then these equations reduce to what we obtained earlier for the circle centered at the origin.

1. If and , prove that the point of tangency (between and ) is precisely the midpoint of the and intercepts of the tangent line. This also generalizes our earlier example.
2. Using the unit circle, explain why the linear-quadratic system

will yield the values of and [consequently, ]. Draw a diagram to begin!

3. Find two numbers whose sum is and whose product is .
4. The difference between two numbers is . The sum of their squares is . Find the two numbers.