Orthogonal Curves: X^2 + 2Y^2 = k and Y = cX^2

[Jakotheshadows]

show that X^2 + 2Y^2 = k and Y = cX^2 are orthogonal trajectories.
I can differentiate these, but I don't know what to do with their derivatives once I find them. I'm stuck :?

 

[soroban]

Re: Orthogonal Curves

Hello, Jakotheshadows!

You need to understand what Orthogonal Trajectories are.
. . They are curves which are perpendicular where they intersect.

Show that \(\displaystyle x^2 + 2y^2 \:=\: k\) and \(\displaystyle y \:=\: cx^2\) are orthogonal trajectories.

You need to:

. . \(\displaystyle \begin{array}{cc}(1) &\text{Find their point(s) of intersection.} \\ (2) & \text{Find the slopes at each intersection.} \\ (3) & \text{Show that the slopes are perpendicular} \\ & \text{at each intersection.} \end{array}\)

 

[BigGlenntheHeavy]

Re: Orthogonal Curves

\(\displaystyle x^{2}+2y^{2} = k \ and\ y = Cx^{2}\)

\(\displaystyle y' =\frac{-x}{2y} \ and \ y' = 2Cx\)

\(\displaystyle (\frac{-x}{2y})(2Cx) = \frac{-2Cx^{2}}{2y} = \frac{-Cx^{2}}{y} \ (y=Cx^{2})\ = \frac{-y}{y} = -1\)

Hence the tangent lines whereever these two equations intersect are perpendicular (orthogonal trajectories).

 

[Deleted member 4993]

Re: Orthogonal Curves

\(\displaystyle x^{2}+2y^{2} = k \ and\ y = Cx^{2}\)

\(\displaystyle y' =\frac{-x}{2y} \ and \ y' = 2Cx\)

\(\displaystyle (\frac{-x}{2y})(2Cx) = \frac{-2Cx^{2}}{2y} = \frac{-Cx^{2}}{y} \ (y=Cx^{2})\ = \frac{-y}{y} = -1\)

Hence the tangent lines whereever these two equations intersect are perpendicular (orthogonal trajectories).

You had solved a similar problem for the student :

viewtopic.php?f=3&t=33840&p=131413#p131413

the student somehow gets the point - and does not get it. If you keep on solving these for him - there is no incentive for him to try.

 

[BigGlenntheHeavy]

Re: Orthogonal Curves

That's his problem, not mine. On the other hand, he might be muleheaded and needs constant reinforcement before he or she gets it.

 

[mmm4444bot]

… That's his problem, not mine …

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