parabola and rectangular hyperbola?
parabola y^2=4ax general point (at^2,2at) lies on he parabola (i dont know if you need this)
find the tangent to the parabola (4a,4a)
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y^2 = 4ax
=> 2y dy/dx = 4a
=> dy/dx = 2a/y
=> dy/dx at (4a, 4a) = 2a/4a = 1/2
=> the equation of the tangent to the parabola y^2 = 4ax at (4a, 4a) is
y - 4a = (1/2) (x - 4a)
=> x - 2y + 4a = 0.