Question

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

Answer 1

Given curve is y² – 2x³ – 4y + 8 = 0

Let the point of contact be (h, k)

Equation of tangent at (h, k) is

y – k = (3h²/(k – 2))(x – h)

which is also passing through (1, 2), so

⇒ 2 – k = (3h²/(k – 2))(1 – h)

⇒ 3h² – 3h³ = –(k² – 4k + 4)

⇒ 3h² – 3h³ + (k² – 4k + 4) = 0 ...(i)

Also, (h, k) lies on the curve, so

k² – 2h³ – 4k + 8 = 0

⇒ k² – 4k + 4 = 2h³ – 4 ...(ii)

From (i) and (ii), we get,

3h² – 3h³ + 2h³ – 4 = 0

⇒ –h³ + 3h² – 4 = 0

⇒ h³ – 3h² + 4 = 0

⇒ (h + 1)(h – 2)² = 0

⇒ h = –1, 2

when h = –1, then k² – 4k + 10 = 0

Thus k is imaginary

So h = –1 is rejected.

When h = 2, then k² – 4k – 8 = 0

Thus, k = 2 ± √3

Therefore, the points of contact are 

(2, 2 + √3) & (2, 2 – √3)

Hence, the equations of tangents are