Editing Analytic geometry (section)

==Finding intersections of geometric objects {{anchor|Intersections}}==
{{main|Intersection (geometry)}}
For two geometric objects P and Q represented by the relations <math>P(x,y)</math> and <math>Q(x,y)</math> the intersection is the collection of all points <math>(x,y)</math> which are in both relations.<ref>While this discussion is limited to the xy-plane, it can easily be extended to higher dimensions.</ref>

For example, <math>P</math> might be the circle with radius 1 and center <math>(0,0)</math>:  <math>P = \{(x,y) | x^2+y^2=1\}</math> and <math>Q</math> might be the circle with radius 1 and center <math>(1,0): Q = \{(x,y) | (x-1)^2+y^2=1\}</math>.  The intersection of these two circles is the collection of points which make both equations true.  Does the point <math>(0,0)</math> make both equations true?   Using <math>(0,0)</math> for <math>(x,y)</math>, the equation for <math>Q</math> becomes <math>(0-1)^2+0^2=1</math> or <math>(-1)^2=1</math> which is true, so <math>(0,0)</math> is in the relation <math>Q</math>.  On the other hand, still using <math>(0,0)</math> for <math>(x,y)</math> the equation for <math>P</math> becomes <math>0^2+0^2=1</math> or <math>0=1</math> which is false.  <math>(0,0)</math> is not in <math>P</math> so it is not in the intersection.

The intersection of <math>P</math> and <math>Q</math> can be found by solving the simultaneous equations:

<math display="block">x^2+y^2 = 1</math>
<math display="block">(x-1)^2+y^2 = 1.</math>

Traditional methods for finding intersections include substitution and elimination.

'''Substitution:'''  Solve the first equation for <math>y</math> in terms of <math>x</math> and then substitute the expression for <math>y</math> into the second equation:

<math display="block">x^2+y^2 = 1</math>
<math display="block">y^2=1-x^2.</math>

We then substitute this value for <math>y^2</math> into the other equation and proceed to solve for <math>x</math>:
<math display="block">(x-1)^2+(1-x^2)=1</math>
<math display="block">x^2 -2x +1 +1 -x^2 =1</math>
<math display="block">-2x = -1</math>
<math display="block">x=1/2.</math>

Next, we place this value of <math>x</math> in either of the original equations and solve for <math>y</math>:

<math display="block">(1/2)^2+y^2 = 1</math>
<math display="block">y^2 =3/4</math>
<math display="block">y = \frac{\pm \sqrt{3}}{2}.</math>

So our intersection has two points:
<math display="block"> \left(1/2,\frac{+ \sqrt{3}}{2}\right) \;\;  \text{and} \;\;  \left(1/2,\frac{-\sqrt{3}}{2}\right). </math>

'''Elimination''':  Add (or subtract) a multiple of one equation to the other equation so that one of the variables is eliminated. For our current example, if we subtract the first equation from the second we get <math>(x-1)^2-x^2=0</math>. The <math>y^2</math> in the first equation is subtracted from the <math>y^2</math> in the second equation leaving no <math>y</math> term. The variable <math>y</math> has been eliminated. We then solve the remaining equation for <math>x</math>, in the same way as in the substitution method:

<math display="block">x^2 -2x +1 -x^2 =0 </math>
<math display="block">-2x = -1</math>
<math display="block">x=1/2.</math>

We then place this value of <math>x</math> in either of the original equations and solve for <math>y</math>:
<math display="block">(1/2)^2+y^2 = 1</math>
<math display="block">y^2 = 3/4</math>
<math display="block">y = \frac{\pm \sqrt{3}}{2}.</math>

So our intersection has two points:
<math display="block"> \left(1/2,\frac{+ \sqrt{3}}{2}\right) \;\;  \text{and} \;\; \left(1/2,\frac{-\sqrt{3}}{2}\right). </math>

For conic sections, as many as 4 points might be in the intersection.

===Finding intercepts===
{{main|x-intercept|y-intercept}}
One type of intersection which is widely studied is the intersection of a geometric object with the <math>x</math> and <math>y</math> coordinate axes.

The intersection of a geometric object and the <math>y</math>-axis is called the <math>y</math>-intercept of the object.
The intersection of a geometric object and the <math>x</math>-axis is called the <math>x</math>-intercept of the object.

For the line <math>y=mx+b</math>, the parameter <math>b</math> specifies the point where the line crosses the <math>y</math> axis. Depending on the context, either <math>b</math> or the point <math>(0,b)</math> is called the <math>y</math>-intercept.