# Shortest distance between point and line.

1. Feb 21, 2013

### ninjohn

1. Shortest distance between: 5x^2 - 6xy + 5y^2 = 4 and origin.

2. d = sqrt(x^2 + y^2)

3. d = sqrt[(4+6xy)/5]. Can't figure out how to get an explicit equation in x or y.

This is a first semester calculus problem out of Thomas from my 1971 class. This is indepent study.

Last edited: Feb 21, 2013
2. Feb 21, 2013

### Staff: Mentor

Is this your work for this problem? What you have is so terse, it's difficult to tell what you're doing. Please tell us how you got the above.
Also, when you start a thread, don't delete the three parts in the homework template. They are there for a reason.

3. Feb 21, 2013

### SqueeSpleen

He probably did:
Distance=
$\sqrt{x^2+y^2}$
And
$5x^2-6xy+5y^2=4$
$5x^2+5y^2=4+6xy$
$x^2+y^2=(4+6xy)/5$
ninjohn. How do you generally calculate the minimum distance between a point and a fuction?

4. Feb 21, 2013

### micromass

We prefer it that the OP says how he did it. Of course Mark44 knows how to derive the formula. But the point is that the OP should show his work. For all we know, he did something very wrong. Or his assignment was "prove that $d=\sqrt{(4+5xy)/5}$. In the latter case, you actually gave him the answer!

5. Feb 21, 2013

### ninjohn

Thanks for comments. This is my first use of forum, so please forgive my criptic description or any etiquette faux pas. Both SS & mm have shown work I did to get d^2 = x^2 + y^2 for distance between origin and any (x,y). Rearranging curve equation produces relationship shown above. I would like to take deravitive of 'd' = sqrt[(4+6xy)/5] set = 0 and solve for x and y to get min value. Problem is that last equation has both x & y. Since this is a first semester calc problem, there are no partial derivatives yet. I need another equation to get d=f(x) or d=g(y) explicitely.

6. Feb 21, 2013

### Ray Vickson

Since you have not yet taken Lagrange multiplier methods, you will be forced to do this problem the hard way. You need to reduce everything to a single-variable problem, which you can do by using the constraint equation 5x^2−6xy+5y^2=4 to solve for y as a function of x, say. Of course, there will be two roots, because the quadratic formula has '±' in it. This is not mysterious; it just corresponds to something like an upper and lower branches of an ellipse.

Anyway, since you have (on each branch of the solution) an expression of the form y = h(x), the square of the distance d^2 = x^2 + (h(x))^2 is now a single-variable function.

7. Feb 21, 2013

### SqueeSpleen

Do you know how to reduce a conic to it's canonical form? (I tried to search the book but I didn't find it).