## Guest

Arun
25750 Points
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)