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what is the differential equation of thetangents of the parabola y=x^2?

Bonjoe Tom Isaiah , 9 Years ago
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Vikas TU

Last Activity: 8 Years ago

differential equation of thetangents of the  parabola y=x^2
is given by : 
y = mx – am^2
a =1/4
y = mx -m^2/4................................(1)
Since the tangent eqn. has only one constant
differentiate ir w.r.t x one time and then substitute,
dy/dx = m ------------------------(2)

in eqn. (1)
we get,
y = xdy/dx – (dy/dx)^2/4 is required differential eqn.

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