# Find the point(s) on the parabola x= y^2-8y+18 closest to (-2,4)

1. May 5, 2013

### physphys

1. The problem statement, all variables and given/known data

Find the point(s) on the parabola x= y^2-8y+18 closest to (-2,4)

2. Relevant equations

3. The attempt at a solution
Using the distance formula I found that d^2 = y^4 - 16y^3 + 105y^2 - 328y + 416
after differentiating I have 2d' = 4y^3 - 48y^2 + 210y -328
I'm not sure what to do now, or what I may have done wrong, any help is appreciated.
Thanks!

2. May 5, 2013

### Unredeemed

I'd have used the fact that the point closest to (-2,4) will lie on the line that passes through (-2,4) and is perpendicular to the parabola at the point where they cross.

3. May 5, 2013

### physphys

I don't understand what you mean. We were just taught to use the distance formula D^2 = (X2-X1)^2 + (Y2-Y1)2
I subbed in -2 as x1 and 4 as y1 then the equation of x= y^2 -8y + 18 as X2. Am I doing it wrong?

4. May 5, 2013

### Unredeemed

I feel like using the distance formula is over complicating things. After all, you haven't been asked to find the ACTUAL distance, just the point which is closest. You could do that by minimising the distance, but I think it's easier to think about what this would look like on the graph.

So, the closest point on the parabola is the point which lines on the line which passes through (-2, 4) and intersects the parabola at a right angle.

5. May 5, 2013

### voko

If you have studied Lagrange's multipliers, that might be the easiest solution.

Otherwise, stick with Unredeemed's advice about the geometrical consideration: minimizing the metric formula yields a cubic equation, as you have already learnt.

6. May 5, 2013

### physphys

Unredeemed, can you please show me how to go on with the process you are talking about? We were taught to find the distance formula then differentiate and solve when it is 0. I've never studies Lagrange's multipliers.

7. May 5, 2013

### Alcubierre

Voko, what would the constraint equation be if you applied the method of Lagrange multipliers?

8. May 6, 2013

### Ray Vickson

So far, so good. Now you just need to solve a cubic equation in y. You could try using the Rational Root Theorem, or try to solve the equation numerically, or use the exact formula for the roots of a cubic.

9. May 6, 2013

### HallsofIvy

Staff Emeritus
You don't have to use any Calculus for this particular problem. Completing the square, [tex]x= y^2-8y+18= y^2- 8y+ 16+ 2= (y- 4)^2+ 2. Notice what the vertex is.

10. May 6, 2013

### voko

The equation of the parabola. Minimizing the square of the distance to the point given.

11. May 6, 2013

### ehild

Follow HallsofIvy's hint, write the equation of the parabola in the form x=(y-4)^2+2, by completing the square. Write up d^2 with that formula without expanding it and differentiate. You can also plot out the function and the point and see the solution at once

ehild