Arun
Last Activity: 5 Years ago
For the parabola y' = (1/2) x This is the slope m so
m = (1/2) xp where the subscript p denotes parabola. This implies xp = 2 m
The point of tangency is thus ( 2m, m2 ) where I used yp = (1/4) (xp)2 = m2
Using the point slope formula, the tangent line equation is y = m x - m2 Thus the b value is - m2
For the circle the procedure is the same but more complicated. I start by looking for a point of tangency in the fourth quadrant. There y = - sqrt(4 - x2) and y' = x/Sqrt(4 - x2)
m = xc /sqrt(4 - xc2 ) where the subscript c denotes circle.
This can be rearranged to get xc = 2 m /sqrt( 1 + m2 ) . Proceeding as for the parabola case , the point of tangency is (2m/sqrt(1 + m2) , -2/sqrt(1 + m2)
After some algebra the tangent line formula is y = m x - [ 2 + 2 m2 ]/sqrt(1 + m2)
So the b value is -[2 + 2 m2] / sqrt(1 + m2)
Now the b values for the parabola and the circle must agree. This leads to the equation
m2 = (2 + 2m2 ) / sqrt( 1 + m2 )
This is a tricky equation, but it can be solved exactly, the result is
m = ± sqrt[ 2 + (1/2) sqrt(32) ] ~ ± 2.197
From this it can be seen that there are two common tangents.
y = 2.197 x - (2.197)2 and y = - 2.197 x - (2.197)2
These intersect at the point P = ( 0, - (2.197)2 )
The distance from P to the origin is (2.197)2 = 4.828
The exact value is: 2 + (1/2) sqrt(32)