Question #223874

Find the equation of the family of all erhogonal trajectories of the family ofcircles which pass through origin and hav centres on the y-axis.

Expert's answer

Upon differentiation with respect to x, we obtain

2x+2yy′+2g(1−y′)=0,

where y′ denotes the derivative of y with respect to x. Upon dividing out by 2, we arrive at

x+yy′+g(1−y′)=0,

from which we get

g=x+yy′y′−1.

Putting this value of g into the equation of the family of circles, we get

x2+y2+2x+yy′y′−1(x−y)−2=0,

so

(x2+y2−2)(y′−1)+2(x+yy′)(x−y)=0

or

(x2+y2−2+2xy−2y2)y′+(2x2−2xy−x2−y2+2)=0

or

(x2+2xy−y2−2)y′+(x2−2xy−y2+2)=0,

from which we get

y′=−"\\dfrac{x\u00b2\u22122xy\u2212y\u00b2+2}{x\u00b2+2xy\u2212y\u00b2\u22122}"

Now for the orthogonal trajectories, we get

"y\u2032=\\dfrac{x\u00b2+2xy\u2212y\u00b2\u22122}{x\u00b2\u22122xy\u2212y\u00b2+2}"

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