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I am asked to show that the given families of curves are orthogonal trajectories of each other.

$$x^2+y^2=ax$$ $$x^2+y^2=by$$

I know that two functions are called orthogonal if at every point their tangents lines are perpendicular to each other. If I differentiate both of these functions, and the resulting expressions are reciprocals of one another, I have shown that they are orthogonal trajectories of each other.

1. $$x^2+y^2=ax$$ $$2x+2yy'=a'x+x'a$$ $$y'=\frac{a-x}{2y}$$

  1. $$x^2+y^2=by$$ $$2x+2yy'=b'y+y'b$$ $$y'=\frac{b-2x}{y}$$

The results I get from differentiating these two functions don't seem to be reciprocals of each other. I am wondering if I have differentiated these two functions incorrectly, or if there is a point substitution that will show these two are reciprocals.

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First, you want the slopes to be negative reciprocals. Second, you only need this to happen where they intersect. – Gerry Myerson Jul 9 '12 at 7:46
up vote 5 down vote accepted

Suppose your two curves are defined by $x^2+y^2=ax$ and $x^2+y^2=by$ , with $a,b$ real constants. To compare the tangent line slopes at given points for each curve, we differentiate the first equation to find that $$ 2x + 2yy' = a~~\text{and so}~~y' = \frac{a-2x}{2y}~~, $$ and the second to find that $$ 2x + 2yy' = by' ~~\text{and so}~~ y' = \frac{2x}{b-2y}~~. $$ These answers are different from the ones you got -- note that what I did was group terms containing $y'$ on one side of the equation, and divide through by whatever factor accompanied it.

We are in business so long as the product of these two quantities, for a given pair $(x,y)$ where the curves intersect, is $-1$. So, multiply one by the other and we get $$ \frac{2x(a-2x)}{2y(b-2y)} = \frac{x(a-2x)}{y(b-2y)} ~~. $$ I believe the problem you ran into is that it is not clear (in an algebraic sense anyways) that these factors should cancel in any way. But remember -- since we are looking at a point where the two curves we were given intersect, we may apply both of those equations. Where does $a-2x$ appear?

Well we have $x^2 + y^2 = ax$ , so that $ax - x^2 = y^2$ , $ax - 2x^2 = y^2 - x^2$ , and $x(a-2x) = y^2-x^2$. Now bear with me, while this may not look simpler, observe that by the same token, $$ x^2 + y^2 = by ~~\text{means that}~~ by-2y^2 = x^2-y^2 ~~\text{and so}~~ y(b-2y) = x^2-y^2 ~~. $$ The factors on top and bottom are indeed the same aside for a sign switch, so the two slopes are negative reciprocal.

BTW: A helpful, and even pretty exercise is to actually plot some of these orthogonal curves. In this case, you'll notice that the first family is circles with an $x$-offset of the center from the origin, while the second family is circles with a $y$-offset. Is it clear why these are orthogonal?

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Thank you, this is a great solution. The algebra at the end is very helpful. I see that two functions are orthogonal when, at a given point, their slopes are negative reciprocals of each other. – Kurt Jul 10 '12 at 3:56
Specifically, at all the points where the curves that the functions define intersect. – Eugene Shvarts Jul 10 '12 at 21:06

Start from the first family of curves like this: $$x^2+y^2=ax \longrightarrow 2x+2yy'=a$$ $$\longrightarrow y'=\frac{a-2x}{2y}$$ we see that $a=\frac{x^2+y^2}{x}$ so put it to the last result above. we get:$$y'=\frac{y^2-x^2}{2xy}$$ Now if you want to find orthogonal trajectories of the first family you should solve: $$y'=-\frac{2xy}{y^2-x^2}$$ which is homogenous equation. Solve it and you will find the second family of curves with choosing suitable constant $b$ in the last solution.

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I like it 8-) +1 – amWhy Mar 11 '13 at 0:06

generally two equation of curves are perpendicular to each other,if product of their slopes is $-1$. from first we have $2*x+2*y*y'=a$ so $2*y*y'=a-2*x$ finally we have $y'=(a-2*x)/(2*y)$ from second curve

$2*x+2*y*y'=b*y'$ or $2*x=b*y'-2*y*y'$.

factor out y',we will have $2*x=y'(b-2*y)$ and $y'=(2*x)/(b-2*y)$ if they are orthogonal,it depend what points $(x,y)$ we have,simple you need points to put and see if product of slopes is $-1$

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