# Conic sections

#### MarkFL

Staff member
Conic sections are the curves that are formed when a plane intersects the surface of a right circular cone. These curves are the circle, ellipse, hyperbola, and the parabola. The study of conic sections dates back over 2000 years to ancient Greece. Apollonius of Perga (262-190 B.C.) wrote an eight volume treatise on the subject.

For a proof that the above definition of the conic section is equivalent to the focus-directrix definition we will use, see A Calculus Notebook by C. Stanley Ogilvey (Boston MA.: Prindle, Weber & Schmidt, 1968). The very readable proof there requires only a knowledge of high school geometry; no calculus is used in the presentation.

In the dictionary, we find:

conic section-1: a plane curve, line, or point that is the intersection of or bounds the intersection of a plane and a cone with two nappes. 2: a curve generated by a point which always moves so that the ratio of its distance from a fixed point to its distance from a fixed line is constant.

We'll be using the second definition to obtain the equation for the conic section. The moving point is called the generatrix. The fixed point is the focus and the fixed line in the directrix.

If we define the focus as $\displaystyle (x_0,y_0)$ and the directrix as $\displaystyle y=mx+b$, we then obtain by definition:

$\displaystyle \frac{\sqrt{(x-x_0)^2+(y-y_0)^2}}{\frac{|mx+b-y|}{\sqrt{m^2+1}}}=\frac{\sqrt{(m^2+1)\left((x-x_0)^2+(y-y_0)^2 \right)}}{|mx+b-y|}=e$ where $\displaystyle 0<e$

This constant of proportionality e is referred to as the eccentricity of the conic section. The line perpendicular to the directrix and passing through the focus is called the axis of symmetry. Hence, using the fact that perpendicular slopes have a product of -1, we may use the point-slope formula to find the axis of symmetry:

$\displaystyle y-y_0=-\frac{1}{m}(x-x_0)$

$\displaystyle y=-\frac{x}{m}+\frac{x_0}{m}+y_0$

To find the vertices (sing. vertex), those points where the generatrix crosses the axis of symmetry, we may solve the system:

$\displaystyle y=-\frac{x}{m}+\frac{x_0}{m}+y_0$

$\displaystyle (x-x_0)^2+(y-y_0)^2=\frac{(mx+b-y)^2}{m^2+1}e^2$

Substituting for y from the first equation into the second, we obtain:

$\displaystyle (x-x_0)^2+\left(\left(-\frac{x}{m}+\frac{x_0}{m}+y_0 \right)-y_0 \right)^2=\frac{e^2}{m^2+1}\left(mx+b-\left(-\frac{x}{m}+\frac{x_0}{m}+y_0 \right) \right)^2$

We may simplify as follows:

$\displaystyle (x-x_0)^2+\left(\frac{x-x_0}{m} \right)^2=\frac{e^2}{m^2+1}\left(mx+\frac{x}{m}-mx_0-\frac{x_0}{m}+mx_0+b-y_0 \right)^2$

Notice we added zero on the right in the form $\displaystyle 0=mx_0-mx_0$ to enable us to write:

$\displaystyle \left(1+\frac{1}{m^2} \right)(x-x_0)^2=\frac{e^2}{m^2+1}\left(\left(m+\frac{1}{m} \right)(x-x_0)+(mx_0+b-y_0) \right)^2$

$\displaystyle (x-x_0)^2=\frac{e^2m^2}{(m^2+1)^2}\left(\left(\frac{m^2+1}{m} \right)(x-x_0)+(mx_0+b-y_0) \right)^2$

$\displaystyle (x-x_0)^2=\left(e\left((x-x_0)+\left(\frac{m}{m^2+1} \right)(mx_0+b-y_0) \right) \right)^2$

$\displaystyle x-x_0=\pm e\left((x-x_0)+\left(\frac{m}{m^2+1} \right)(mx_0+b-y_0) \right)$

$\displaystyle (x-x_0)\mp e(x-x_0)=\pm e\left(\frac{m}{m^2+1} \right)(mx_0+b-y_0)$

$\displaystyle (x-x_0)(1\mp e)=\pm e\left(\frac{m}{m^2+1} \right)(mx_0+b-y_0)$

$\displaystyle x-x_0=\pm\frac{e}{1\mp e}\left(\frac{m}{m^2+1} \right)(mx_0+b-y_0)$

$\displaystyle x=x_0\pm\frac{me(mx_0+b-y_0)}{(1\mp e)(m^2+1)}$

Now, to find the y-coordinates of the vertices, we may use:

$\displaystyle y=-\frac{x_0\pm\frac{me(mx_0+b-y_0)}{(1\mp e)(m^2+1)}}{m}+\frac{x_0}{m}+y_0$

$\displaystyle y=y_0\mp\frac{e(mx_0+b-y_0)}{(1\mp e)(m^2+1)}$

Thus, the coordinates of the vertices are:

$\displaystyle \left(x_0\pm\frac{me(mx_0+b-y_0)}{(1\mp e)(m^2+1)},y_0\mp\frac{e(mx_0+b-y_0)}{(1\mp e)(m^2+1)} \right)$

If the eccentricity e is 1, then we obviously have only 1 vertex. We know it lies along the axis of symmetry and must be halfway between the focus and the directrix, since when e = 1 this means the generatrix must be equidistant to the directrix and focus.

In the next installment, we will look at the distance between vertices for those conic section whose eccentricity is not 1.

Comments and questions should be posted here:

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#### MarkFL

Staff member
Okay, we have the two vertices (where $\displaystyle e\ne1$), so let's compute the distance between them. We will label this distance $\displaystyle 2a$. We use this label so that the distance from the point midway between the vertices and either vertex may be labeled $\displaystyle a$.

$\displaystyle 2a=\sqrt{ \begin{array}{ll} \left(\left(x_0+\frac{me(mx_0+b-y_0)}{(1-e)(m^2+1)} \right)\right.&-\left.\left(x_0-\frac{me(mx_0+b-y_0)}{(1+e)(m^2+1)} \right) \right)^2\\ &+\left(\left(y_0-\frac{e(mx_0+b-y_0)}{(1-e)(m^2+1)} \right)-\left(y_0+\frac{e(mx_0+b-y_0)}{(1+e)(m^2+1)} \right) \right)^2 \end{array} }$

$\displaystyle 2a=\sqrt{\left(\frac{2me(mx_0+b-y_0)}{(1-e^2)(m^2+1)} \right)^2+\left(\frac{2e(mx_0+b-y_0)}{(1-e^2)(m^2+1)} \right)^2}$

$\displaystyle 2a=\frac{2e|mx_0+b-y_0|}{\sqrt{m^2+1}|1-e^2|}$

Now, if we observe that the focal parameter $\displaystyle p$, the distance from the focus to the directrix, is:

$\displaystyle p=\frac{|mx_0+b-y_0|}{\sqrt{m^2+1}}$ then we may write:

$\displaystyle a=\frac{ep}{|1-e^2|}$

An eccentricity of 1 seems to be a boundary between two types of conics which have two vertices. When $\displaystyle e<1$, the two vertices must be on the same side of the directrix, since the generatrix must always be closer to the focus, and so it may fall either in between focus and directrix or on the other side of the focus, away from the directrix, forming a closed curve. When $\displaystyle e>1$, the generatrix must be closer to the directrix, thus it may only cross the axis of symmetry once, meaning we must have two open branches.

We may notice that $\displaystyle a(e)=a\left(\frac{1}{e} \right)$. We may be interested in finding where, if any, extrema occur.

$\displaystyle \frac{da}{de}=\frac{p|1-e^2|+ep\frac{2e(1-e^2)}{|1-e^2|}}{(1-e^2)^2}=\frac{p(1-e^2)(1+e^2)}{|1-e^2|^3}$

So, we find no relative extrema, and we note that:

$\displaystyle a(0)=0$

$\displaystyle \lim_{e\to1}a(e)=+\infty$

$\displaystyle \lim_{e\to\infty}a(e)=0$

Back to the definition of conic sections. Recall that our equation from the definition is equivalent to:

$\displaystyle (m^2+1)\left((x-x_0)^2+(y-y_0)^2 \right)=e^2(mx+b-y)^2$

Expand and regroup on like x and y terms:

$\displaystyle (m^2+1)(x^2-2x_0x+x_0^2+y^2-2y_0y+y_0^2)-e^2(m^2x^2+b^2-2mxy+2bmx-2by+y^2)=0$

$\displaystyle c_1x^2+c_2xy+c_3y^2+c_4x+c_5y+c_6=0$

where:

$\displaystyle c_1=m^2(1-e^2)+1$

$\displaystyle c_2=2e^2m$

$\displaystyle c_3=m^2-e^2+1$

$\displaystyle c_4=-2(x_0(m^2+1)+be^2m)$

$\displaystyle c_5=2(be^2-y_0(m^2+1))$

$\displaystyle c_6=(x_0^2+y_0^2)(m^2+1)-(be)^2$

This is the general equation of the conic section in terms of its focus, directrix and eccentricity.

Next week, we will look at the discriminant of the equation, to see what this tells us about the type of conic section we have.

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#### MarkFL

Staff member
As we have found, an equation of the form:

$\displaystyle Ax^2+Bxy+Cy^2+Dx+Ey+F=0$

is that of a conic section. The discriminant $\displaystyle \Delta=B^2-4AC$ can provide information about the type of conic section it is, except in degenerate cases. The graph is:

i) a parabola if $\displaystyle \Delta=0$.

ii) an ellipse if $\displaystyle \Delta<0$.

iii) a hyperbola if $\displaystyle \Delta>0$.

The discriminant of the conic section we derived above is:

$\displaystyle \Delta=(2e^2m)^2-4(m^2(1-e^2)+1)(m^2-e^2+1)=4(e^2-1)(m^2+1)^2$

Hence, we see that we have:

i) a parabola if $\displaystyle e=1$.

ii) an ellipse if $\displaystyle e<1$.

iii) a hyperbola if $\displaystyle e>1$.

This agrees with our findings above. Let's now look at the different types of conic sections in more detail.

circle - a closed plane curve every point of which is equidistant from a fixed point (center or focus) within the curve.

Suppose we define the focus to be the point $\displaystyle (h,k)$ in the plane. Since a circle is the set of all points equidistant from the focus (we'll let this constant distance be $\displaystyle r$), we may use the formula for distance to obtain the equation:

$\displaystyle \sqrt{(x-h)^2+(y-k)^2}=r$

Squaring, we find:

$\displaystyle (x-h)^2+(y-k)^2=r^2$

This of course is the equation of the circle having radius $\displaystyle r$ and center $\displaystyle (h,k)$.

Recall our general conic equation. What condition(s) are necessary to transform it into a circular equation? Note the circular equation has no $\displaystyle xy$ term, thus either $\displaystyle e$ or $\displaystyle m$ must be zero. Since we require $\displaystyle e>0$ (otherwise we have a degenerate conic, a circle of zero radius, i.e., a point) let's look at $\displaystyle e$ approaching zero. Recall that from the definition of the conic section we obtained the equation:

$\displaystyle (x-x_0)^2+(y-y_0)^2=\frac{e^2(mx+b-y)^2}{m^2+1}$

The point on the curve closest to the directrix is the vertex (a point where the axis of symmetry of a conic section intersects the curve itself). Recall also that the focal parameter $\displaystyle p$ is the distance from the focus to the directrix, and so the distance $\displaystyle r$ from the focus to the vertex is:

$\displaystyle r=\frac{1}{\frac{1}{e}+1}p=\frac{ep}{e+1}\, \therefore\,p=\frac{r}{e}(e+1)$

If we hold this value $\displaystyle r$ constant while the eccentricity $\displaystyle e$ approaches zero, we see that the focal parameter must approach infinity. Therefore, the $\displaystyle y$-intercept of the directrix would be a funtion of $\displaystyle e$. We know that the focal parameter $\displaystyle p$ is given by:

$\displaystyle p=\frac{|mx_0+b-y_0|}{\sqrt{m^2+1}}$

Solving this for $\displaystyle b$, we find:

$\displaystyle b=(y_0-mx_0)\pm p\sqrt{m^2+1}=(y_0-mx_0)\pm \frac{r}{e}(e+1)\sqrt{m^2+1}$

Now we are ready to evaluate our limit.

$\displaystyle \lim_{e\to0}\left[\frac{e^2\left(mx+\left((y_0-mx_0)\pm \frac{r}{e}(e+1)\sqrt{m^2+1} \right)-y \right)^2}{m^2+1} \right]$

Simplifying, we find:

$\displaystyle \lim_{e\to0}\left[\frac{(emx+e(y_0-mx_0)\pm r(e+1)\sqrt{m^2+1}-ey)^2}{m^2+1} \right]=r^2$

Hence, we have shown that as the eccentricity approaches zero and the distance from the focus to vertex $\displaystyle r$ is held constant, the directrix "recedes into infinity" and the conic equation is given by:

$\displaystyle (x-x_0)^2+(y-y_0)^2=r^2$

This, as we saw earlier, is the set of all points equidistant from the focus (which is the center of the circle), where this constant distance is $\displaystyle r$, the radius of the circle.

Next week, we shall look at the ellipse.

#### MarkFL

Staff member
The Ellipse

ellipse
- a closed plane curve generated by a point moving in such a way that the sums of its distances from two fixed points (foci) is a constant: a plane section of a right circular cone that is a closed curve.

Terminology for the ellipse:

1. The focal axis is the line passing through the foci.

2. The center is the point midway between the foci, on the focal axis.

3. The focal distance is the distance from the center to either focus.

4. The vertices are the two points where the focal axis meets the ellipse.

5. The major axis is the line segment joining the vertices.

6. The minor axis is the line segment through the center of the ellipse, perpendicular to the major axis and with endpoints on the ellipse.

7. The eccentricity is the ratio of the focal distance to half the length of the major axis.

There is a simple mechanical method for constructing an ellipse that arises directly from the first definition of the curve given above. Mark the foci, $F_1(x_1,y_1)$ and $F_2(x_2,y_2)$ on a drawing board and insert thumbtacks at those points. Now take a piece of string whose length is greater than the distance between the foci and attach the ends of the string to the tacks. Next, pull the string taut with a pencil tip, then move the pencil, tracing a curve as you keep the string taut. The curve you draw will be elliptical. As you can see, the sum of the distance from the pencil tip to the two tacks is kept constant.

This should be enough to convince the reader that the ellipse will enclose the foci. We may state this definition mathematically as follows:

$\sqrt{(x-x_1)^2+(y-y_1)^2}+\sqrt{(x-x_2)^2+(y-y_2)^2}=S$

where $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}<S$.

The equation of the line passing through the foci (the focal axis) can be found by the point-slope formula (assuming it is not vertical):

$\displaystyle y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1$

or equivalently:

$\displaystyle y=\frac{y_2-y_1}{x_2-x_1}(x-x_2)+y_2$

Now, to find the vertices, we may solve the above non-linear system. Substituting for $y$ from the linear equation into that of the ellipse, we find:

$\displaystyle \sqrt{(x-x_1)^2+\left(\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1-y_1 \right)^2}+\sqrt{(x-x_2)^2+\left(\frac{y_2-y_1}{x_2-x_1}(x-x_2)+y_2-y_2 \right)^2}=S$

$\displaystyle |x-x_1|\sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}+|x-x_2|\sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}=S$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(|x-x_1|+|x-x_2| \right)=S$

Let's require $x_1<x_2$ and examine the following cases:

i) $x_1\le x\le x_2$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(x-x_1+x_2-x \right)=S$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(x_2-x_1 \right)=S$

$\displaystyle \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=S$

Recall this violates a stated requirement. This leads to a degenerate ellipse, so we know no vertices will lie within this range.

ii) $x<x_1<x_2$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(x_1-x+x_2-x \right)=S$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(x_1+x_2-2x \right)=S$

$\displaystyle x_1+x_2-2x=\frac{S}{\sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}}$

$\displaystyle x=\frac{x_1+x_2}{2}-\frac{S}{2\sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}}$

and so:

$\displaystyle y=\frac{y_1+y_2}{2}-\frac{S(y_2-y_1)}{2(x_2-x_1)\sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}}$

iii) $x_1<x_2<x$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(x-x_1+x-x_2 \right)=S$

Similar to the technique used in the second case, we find:

$\displaystyle x=\frac{x_1+x_2}{2}+\frac{S}{2\sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}}$

$\displaystyle y=\frac{y_1+y_2}{2}+\frac{S(y_2-y_1)}{2(x_2-x_1)\sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}}$

Next week we shall look at defining the major and minor axes.

#### MarkFL

Staff member
The major axis of an ellipse, usually denoted by $a$ is the distance from the mid-point of the foci to the vertices. For convenience, let's define $F$ to be half the distance between foci, and the vertices are then given by:

$\displaystyle \left(\frac{x_1+x_2}{2}\pm\frac{S(x_2-x_1)}{F},\frac{y_1+y_2}{2}\pm\frac{S(y_2-y_1)}{F} \right)$

Hence:

$\displaystyle a=\frac{S}{2}$

as we should expect.

The equation of the line passing through the mid-point of the foci but perpendicular to the focal axis is:

$\displaystyle y=\frac{x_2-x_1}{y_1-y_2}\left(x-\frac{x_1+x_2}{2} \right)+\frac{y_1+y_2}{2}$

The distance from the mid-point of the foci to the point where the ellipse crosses this line is called the minor axis (denoted by $b$), and its length may be found from the right triangle formed by the line segment (minor axis), the line segment from the focal mid-point to a focus and the line segment from the focus to the point where the ellipse touches the minor axis. There results:

$F^2+b^2=\left(\frac{S}{2} \right)^2=a^2$

$b=\sqrt{a^2-F^2}$

Now, to greatly simplify matters, we shall orient our coordinate axes such that the $x$-axis is the focal axis and the origin coincides with the focal mid-point. The equation from the definition of the ellipse is now:

$\displaystyle \sqrt{(x-(-F))^2+(y-0)^2}+\sqrt{(x-(F))^2+(y-0)^2}=2a$

$\displaystyle \sqrt{(x+F)^2+y^2}=2a-\sqrt{(x-F)^2+y^2}$

Square through:

$\displaystyle (x+F)^2+y^2=4a^2-4a\sqrt{(x-F)^2+y^2}+(x-F)^2+y^2$

Expand and simplify:

$\displaystyle \frac{Fx-a^2}{a}=\sqrt{(x-F)^2+y^2}$

Square through again:

$\displaystyle a^2-2Fx+\left(\frac{F}{a}x \right)^2=x^2-2Fx+F^2+y^2$

$\displaystyle a^2+\frac{F^2}{a^2}x^2=x^2+y^2+F^2$

Recall $F^2=a^2-b^2$:

$\displaystyle a^2+\frac{a^2-b^2}{a^2}x^2=x^2+y^2+a^2-b^2$

$\displaystyle \frac{b^2}{a^2}x^2+y^2=b^2$

$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

This is the equation of the ellipse centered at the origin, with major axis $a$ and minor axis $b$. The equation of the ellipse centered at the origin and having a vertical major axis would of course be:

$\displaystyle \frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

If one is presented with two foci which are not centered on the origin, then a simple translation of axes may be used to get:

$\displaystyle \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

where $(h,k)$ is the focal mid-point.

Now, we may also approach this by going back to:

$\displaystyle c_1x^2+c_2xy+c_3y^2+c_4x+c_5y+c_6=0$

It we let the directrix be the horizontal line $y=k$, let the axis of symmetry by the $y$-axis, and a focus at $(0,y_0)$, midpoint of vertices at the origin, then we have:

$x^2+(1-e^2)y^2+2(e^2k-y_0)y+y_0^2-(ek)^2=0$

and our two vertices must occur at:

$\displaystyle \left(0,\frac{y_0\pm ek}{1\pm e} \right)$

For these vertices to have the origin as their mid-point, this implies:

$y_0=e^2k$

and so the vertices are:

$(0,\pm ek)$

and the conic equation then reduces to:

$\displaystyle \frac{x^2}{y_0(k-y_0)}+\frac{y^2}{ky_0}=1$

Recall that for an ellipse, we have $\displaystyle e=\frac{F}{a}$ (assuming the word eccentricity is used to mean the same thing). This means:

$\displaystyle F=e^2k=y_0$

Recall that we found:

$b=\sqrt{a^2-F^2}$

We have from this:

$b=\sqrt{(ek)^2-y_0^2}=\sqrt{y_0(k-y_0}$

$a=\sqrt{y_0(k-y_0)+y_0^2}=\sqrt{y_0k}$

And thus the conic equation may be written:

$\displaystyle \frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

This, as we have already established, is the equation of an ellipse centered at the origin, with a vertical focal/major axis, i.e., axis of symmetry.

We may now also define the ellipse to be a curve generated by a point moving such that the ratio of its distance from a fixed point to its distance form a fixed line is a constant greater than zero but less than one.

Next week we will explore the hyperbola.

#### MarkFL

Staff member
The Hyperbola

hyperbola
-a plane curve generated by a point so moving that the difference of the distances from two fixed points is a constant : a curve formed by the intersection of a double right-circular cone with a plane that cuts both halves of the cone.

Terminology for the hyperbola:

1. The focal axis is the line passing through the foci.

2. The center is the point midway between the foci.

3. The vertices are the two points where the hyperbola meets the focal axis.

4. The transverse axis is the line segment joining the vertices.

5. The conjugate axis is the line segment through the center of the hyperbola, perpendicular to the transverse axis.

6. The eccentricity is the constant ratio for any point $(x,y)$ on the hyperbola of the distance from either focus to the distance from the corresponding directrix.

We may state the first definition above mathematically as follows:

$\sqrt{(x-x_1)^2+(y-y_1)^2}-\sqrt{(x-x_2)^2+(y-y_2)^2}=\pm S$

where $|S|<\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

The plus-or-minus sign indicates that as $S$ is the difference between distances from the generatrix to two different points, we do not require the distance to one focus to always be the greater.

The equation of the line passing through the foci (the focal axis) can be found by the point-slope formula (assuming it is not vertical):

$\displaystyle y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1$

or equivalently:

$\displaystyle y=\frac{y_2-y_1}{x_2-x_1}(x-x_2)+y_2$

Now, to find the vertices, we may solve the above non-linear system. Substituting for $y$ from the linear equation into that of the hyperbola, we find:

$\displaystyle \sqrt{(x-x_1)^2+\left(\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1-y_1 \right)^2}-\sqrt{(x-x_2)^2+\left(\frac{y_2-y_1}{x_2-x_1}(x-x_2)+y_2-y_2 \right)^2}=\pm S$

$\displaystyle |x-x_1|\sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}-|x-x_2|\sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}=\pm S$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(|x-x_1|-|x-x_2| \right)=\pm S$

Let's require $x_1<x_2$ and examine the following cases:

i) $x_1\le x\le x_2$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(x-x_1-x_2+x \right)=\pm S$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(2x-(x_1+x_2) \right)=\pm S$

$\displaystyle x=\frac{x_1+x_2}{2}\pm\frac{S}{2\sqrt{1+\left( \frac{y_2-y_1}{x_2-x_1} \right)^2}}$

and so:

$\displaystyle y=\frac{y_1+y_2}{2}\pm\frac{S(y_2-y_1)}{2(x_2-x_1)\sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}}$

ii) $x<x_1<x_2$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(x_1-x-x_2+x \right)=\pm S$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(x_1-x_2 \right)=\pm S$

$\displaystyle -\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\pm S$

This violates the restriction $|S|<\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$, and so no vertices will be found here.

iii) $x_1<x_2<x$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(x-x_1-x+x_2 \right)=\pm S$

$\displaystyle \sqrt{1+\left(\frac{y_2-y_1}{x_2-x_1} \right)^2}\left(x_2-x_1 \right)=\pm S$

$\displaystyle \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\pm S$

This violates the restriction $|S|<\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$, and so no vertices will be found here.

We may conclude then that the vertices must lie within the rectangle whose diagonal is the line segment joining the foci.

Next week we shall look at defining the major and minor axes.