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


*

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


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. 
 A: 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? 
A: 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.
A: 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 $-1$
