Question

Find the equations of the tangents drawn to the curve y² – 2x³ – 4y + 8 = 0 from point (1, 2).

Answer 1

The equation of the given curve is

y² – 2x³ – 4y + 8 = 0

⇒ dy/dx = 3x²/(y – 2)

Let the point (α, β) lies 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) lies 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 = ± 2√3

Hence, the equation of the tangents are

(y – (2 ± 2√3)) = ± 2√3(x – 2)