y4 - 4x4 - 6xy = 0
Differentiating both sides w.r.t. x, we get
= slope of the tangent at (1, 2)
∴ the equation of normal at M (1, 2) is
The slope of normal at (1, 2)
∴ the equation of normal at M (1, 2) is
Hence, the equations of tangent and normal are 14x – 13y + 12 = 0 and 13x + 14y – 41 = 0 respectively.