The equation of the given curve is
y2 – 2x3 – 4y + 8 = 0
⇒ dy/dx = 3x2/(y – 2)
Let the point (α, β) **** on the curve
Thus, m = (dy/dx)(α, β) = 3α2/(β – 2)
Therefore, the equation of the tangent at
(α, β) is y – β = ( 3α2/(β – 2))(x – α) ...(1)
which is passing through (1, 2)
So, (2 – β) = ( 3α2/(β – 2)) (1 – α) ...(2)
Also, the point (α, β) **** on the curve
y2 – 2x3 – 4y + 8 = 0
So, β2 – 2α2 – 4β + 8 = 0 ...(3)
From (2) and (3), we get,
2(α3 – 2) = 3α2 (α – 1)
⇒ α3 – 3α2 + 4 = 0
⇒ α = 2
when α = 2, β = ±u20092√3
Hence, the equation of the tangents are
(y – (2 ± 2√3)) = ±u20092√3(x – 2)
