Finding Intersections
• Equation of arc 1: y
 
.
0070 x
2 
1 .
2349 x
• Equation of arc 2: y
 
.
0254 x
2 
4 .
4856 x

136 .
9897
• To find intersections, use substitution:
» EQ ARC 1 = EQ ARC 2


.
0070 x
2 
Finding Intersections
1 .
2349 x
 
.
0254 x
2 
4 .
4856 x

136 .
9897

.
0070 x
2 
1 .
2349 x  
.
0070 x
2 
1 .
2349 x
0
 
.
0184 x
2 
3 .
2507 x

136 .
9897
• I solved the system of equations so that one side equaled 0. Now, I will use the quadratic formula to find my x-values.

0

Finding Intersections

.
0184 x
2 
3 .
2507 x

136 .
9897 x x

 x

 b
2 a

3 .
2507


2 (

.
0184 )

3 .
2507

.
0368 b
2


4 ac
2 a
3 .
2507
2
10 .

4 (

.
0184 )(

136 .
9897 )
5671
.
2 (

.
0184 )

10 .
0824
0368

Finding Intersections x x





.
0368
3 .
2507

3
.
.
2507
0368


.
.
.
4846
0368
6961
.
0368 x

88 .
3342

18 .
9168 x

107 .
2510 x

69 .
4174

Finding Intersections
• Now that we have found the x-values of the intersections, substitute into an equation for an arc to find the y-value.
• I find the non-translated one to be just a bit easier.

y

Finding Intersections

.
0070 x
2 
1 .
2349 x x x


107 .
2510
69 .
4174 y
 
.
0070 ( 107 .
2510 )
2 
1 .
2349 ( 107 .
2510 ) y
 
.
0070 ( 11502 .
7770 )

1 .
2349 ( 107 .
2510 ) y y



80 .
5194
51 .
9249

132 .
4443

Finding Intersections x

107 .
2510 y
 
.
0070 x
2 
1 .
2349 x x

69 .
4174 y y

 
.
0070 ( 69 .
4174 )
2

.
0070 ( 4818 .
7754 )

1 .
2349 ( 69 .
4174 )

1 .
2349 ( 69 .
4174 ) y y



33 .
7314
51 .
9921

85 .
7235
I know that the y-values should be exactly the same since I have the same line of symmetry. These are basically the same, where the rounding I have done caused the difference. I’ll label the y-value as the midpoint between the two solutions I got – 59.9585

Finding Intersections