Implicit Differentiation and Orthogonal Trajectories

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Homework Statement

Q 50: The ellipse 3x² +2y² = 5 and y³ = x²
HINT: The curves intersect at (1,1) and (-1,1)
Two families of curves are said to be orthogonal trajectories (of each other) if each member of one family is orthogonal to each member of the other family. Show that the families of curves given in Exercises 51 and 52 are orthogonal trajectories

Exercise 51: The family of circles x² + y² = r² and the family of lines y = mx

Exercise 52: The family of parabolas x = ay² and the family of ellipses x² + (1/2) y² = b

Homework Equations

dy/dx

The Attempt at a Solution

I was a little confused about what this question was asking. This is the only explanation in the book of orthogonal trajectories and the one thing that i know when i see the word orthogonal is that there must be 90 degrees involved. What I assume this question is asking is to prove that each family in Exercises 51 and 52 are orthogonal to each other, basically meaning that their derivatives are perpendicular to each other. dy/dx is the slope of the function and if each group has a dy/dx which is the negative reciprocal of the other group then that means that they are normal to each other which means that they are perpendicular. My approach will be the following, to take the derivative of each family and determine that the derivatives are reciprocals of each other, however i am confused as to what the derivatives of r, a and b will be which is why i assume them to be constants. I also do not understand why the 3x² +2y² = 5 and y³ = x² are given, and I am also assuming that when we have the derivatives, then we plug those back into the function and the values of (1,1) and (-1,1) to prove that the functions intersect each other at those points. Please help me with going through this problem because I'm not really sure what to do and I am sure that my assumptions are not correct for r, a and b being constants. Thank you.
I did take the derivatives with my assumptions:
3x² +2y² = 5
6x + 4y * (dy/dx) = 0
dy/dx = -3x/2y
y³ = x²
3y²*(dy/dx) = 2x
dy/dx = 2x/3y²

51:
y = mx
dy/dx = m(1) = m = rise/run = y/x
x² + y² = r²
2x + 2y * (dy/dx) = 0
dy/dx = -x/y
In this case, assuming that my assumptions are correct, dy/dx for each family are negative reciprocals of each other confirming that they are perpendicular lines, however i do not know how to prove that they interesect at (1,1) and (-1,1)

52:
x = ay²
1 = 2ay * (dy/dx)
dy/dx = 1/2ay
x² + (1/2) y² = b
2x + y * (dy/dx) = 0
dy/dx = -2x/y
Clearly i reach a problem here where they are not related and i assume that either my assumptions are incorrect, i have to use the 3x² +2y² = 5 and y³ = x², or these two families of functions are not orthogonal trajectories of each other.
I'm sure I've done something wrong, any help is greatly appreciated.

 

[Dick]

Q50 has nothing to do with Q51 and Q52. It's just an example. Don't try to apply anything there to the other two problems. You got through Q51 just fine. Now look at Q52, you've got y'=1/(2ay) for one family of curves and y'=(-2x)/y for the other. You CAN use x=ay^2 again to prove they are orthogonal. Try it.

 

[HallsofIvy]

Don't worry about "90 degrees". Two lines, y= mx+ b and y= nx+ c, are "orthogonal" if and only if mn= -1. For curves, two curves, y= f(x) and y= g(x), are "orthogonal" if and only if [itex]f'(x_0)g'(x_0)= -1[/itex] where [itex]x= x_0[/itex] is the x-value at their point of intersection.