# Math Help - Shortest distance between parabola and circle.

1. ## Shortest distance between parabola and circle.

find the shortest distance between parabola y^2=4x and the circle x^2+y^2-24y+128=0

2. Originally Posted by prasum
find the shortest distance between parabola y^2=4x and the circle x^2+y^2-24y+128=0
First draw the graph of each curve to see what's happening.

Are you expected to use calculus to find the answer?

3. i have drawn the graphs but can we do it by solving without calculus

4. Originally Posted by prasum
find the shortest distance between parabola y^2=4x and the circle x^2+y^2-24y+128=0
Originally Posted by prasum
i have drawn the graphs but can we do it by solving without calculus
The shortest distance between parabola and circle is the same as the distance between the center of the circle and one certain point of the parabola, reduced by the radius of the circle.

As far as I understand the question there are 2 different ways, but each requires calculus in the end:

1. A straight line which contains the shortest distance as a segment passes through the center of the circle and is perpendicular to a tangent of the parabola. To determine the slope of the tangent you'll need calculus.

2. Calculate the distance between the center of the circle and any point of the parabola. Determine the shortest distance. For this step you'll need calculus too.

5. Originally Posted by prasum
i have drawn the graphs but can we do it by solving without calculus
Why?

6. we can do it by finding the equation of normal to parabola and the n finding its point of intersection with the parabola.am i right can yuou tell me the calculus method

7. Originally Posted by prasum
find the shortest distance between parabola y^2=4x and the circle x^2+y^2-24y+128=0
There is a non-calculus method, not any simpler though...

You can use the fact that the tangent to the parabola $y^2=4x$ at the point $\left(x_i,\ y_i\right)$

crosses the y-axis at the point $\displaystyle\left(0,\ \frac{y_i}{2}\right)$

You can continue in a few different ways, by finding the intersection of a locus and the parabola,
or the intersection of an ellipse and the parabola.

Alternatively, use similar triangles

$\displaystyle\frac{\left(\frac{y}{2}\right)}{x}=\f rac{x}{12-y}$

$\displaystyle\frac{y}{2x}=\frac{x}{12-y}$

Then, since (x,y) is on the parabola $\displaystyle\Rightarrow\ y^2=4x\Rightarrow\ x=\frac{y^2}{4}$

giving

$\displaystyle\frac{2}{y}=\frac{y^2}{4(12-y)}\Rightarrow\ y^3=8(12-y)$

$y^3+8y-96=0\Rightarrow\ y(y^2+8)=4(24)$

This allows you to find the co-ordinates of the point on the parabola that is closest to the circle.
Find the distance of that point to the circle centre and subtract the radius.

8. At the point $(t^2,2t)$ on the parabola, the gradient of the tangent is 1/t. (You need calculus to justify that, but maybe you are allowed to quote that fact.) So the normal has gradient –t. Thus the equation of the normal is $y-2t = (-t)(x-t^2)$. At the point on that parabola that is closest to the circle, the normal will pass through the centre of the circle. Putting the point (0,12) into the equation of the normal, you get $12-2t = t^3$, and the solution t=2 should jump out at you. (It's a cubic equation, but it only has the one real solution.)

So you get the closest point on the parabola by putting t=2. From there on you can continue as in Archie Meade's solution.