# Minimum Distance between two curves

1. Oct 13, 2014

### ManInTheSuit97

Minimum Distance between y^2=4x and x^2+y^2-12x+31=0.
Attempt:I got that the parabola has vertex at(0,0) and focus at(1,0).The Circle is centred at (6,0)and its radius is sqrt 5.I figured that the double ordinate that passed through (6,0) would be bisected at the point.So I found out the chord of contact and it turned out to be x=6.I substituted the value of x in the parabola and found out y= sqrt24.I thought maybe the minimum distance is the difference in vertical distance and so my answer was sqrt 24-sqrt5.But it is not the answer and so I probably have messed up somewhere.I want to know the approach I should take

2. Oct 13, 2014

### willem2

For any point P outside a circle, the closest point to P on the circle will lie on the line from P to the center of the circle.
The distance From P to the circle will be the distance from P to the center of the circle minus the radius of the circle.
Compute the distance from a point on the parabola to the center of the circle and find a minimum by differentiating this distance.
(it will be easier to find a minimum for the square of the distance)

3. Oct 13, 2014

### RUber

If you are looking to solve this without differentiating, you could first solve the parabola I terms of x, ie $(x,2\sqrt{x})$.
Then write the equation for distance to (6,0).
$d=\sqrt{(x-6)^2+(2\sqrt{x})^2}$
You could use the quadratic equation to find the zero for distance. Then, note that the closest real point to an imaginary number a+bi is just the real part a.