Finding tangent and normal plane equation of curve y ^ 2 = 2mx Z ^ 2 = M-X at point (x.y.z.)

Let y ^ 2 = 2Mt, Z ^ 2 = M-T, x = t, let XYZ be derivative of T, then y '= m / √ 2Mt, Z' = - 1 / √ 2 (M-T) x '= 1, so the tangent vector is (1, M / √ 2mx., - 1 / √ 2 (M-X.))

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