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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

In the dictionary, we find:

We'll be using the second definition to obtain the equation for the conic section. The moving point is called the

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

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

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

To find the

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

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

Substituting for

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

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

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

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

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

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:

http://mathhelpboards.com/commentary-threads-53/commentary-conic-sections-4212.html

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:

http://mathhelpboards.com/commentary-threads-53/commentary-conic-sections-4212.html

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