What is the number of tangent to the curve y^2 - 2x^3 - 4y + 8=0

Question

that pass through (1,2) ? 

Answer 1

The curve is y² - 2x³ - 4y + 8 = 0 and the point is (1, 2).

We can use implicit differentiation to find y '.

2yy ' - 6x² - 4y ' = 0

y '(2y - 4) = 6x²

y ' = (6x²)/(2y - 4).

Substitute the values of (x, y ) = (1, 2) in the above equation.

y ' = (6(1)²)/(2(2) - 4).

y ' = (6)/(4 - 4)

y ' = 6/0.

y ' does not seem to exist at the point (1, 2).

So, we can't find the equation of a tangent line to the curve y² - 2x³ - 4y + 8 = 0 at the point (1, 2).