# Thread: coordinate geometry,parabola , circle

1. ## coordinate geometry,parabola , circle

find the shortest distance between y^2=4x and x^2+y^2-12x+31=0.

2. Originally Posted by Pulock2009
find the shortest distance between y^2=4x and x^2+y^2-12x+31=0.
The firsdt equation describes a parabola opening to the right.

The second equation describes a circle around C(6, 0) with $r = \sqrt{5}$.

Enlarge the circle around C until it touches the parabola. The difference between the larger radius and the original radius is the shortest distance between the parabola and the circle:

$p: y^2=4x$

$c: (x-6)^2+y^2=r^2$ New circle with variable radius.

Point of intersection:

$(x-6)^2+4x=r^2$ Solve for x:

$x=4 \pm \sqrt{r^2-20}$

There is only one point of intersection, that means the 2 curves are tangent, if

$r = \sqrt{20} = 2\sqrt{5}$

The shortest distance is therefore:

$d_{min} = 2\sqrt{5}-\sqrt{5} = \sqrt{5}$

3. Very nicely done!