Find the equation of the line that is tangent to the graph of:
x^2cosy +5y^2=(1/4)x^2 , at the point (2π,π)
x^2(-siny)dy/dx + 2xcosy + 10y dy/dx = (1/2)x
dy/dx = ((1/2x - 2xcosy)/(-siny + 10y)
for the given point (2π,π)
dy/dx = (π - 2π(-1))/(0+10π)
= 3π/(10π) = 3/10
equation:
(y-π) = (3/10)(x-2π)
10y - 10π = 3x - 6π
10y = 3x + 4π
y = (3/10)x + 4π/10