Find the points on the curve (points of tangency)

Question

iven the equation of the curve and the slope of the tangent line? 

Equation of the curve : y^3 + x^2 = y^2 - 5y + 14 
Slope : 1 

Im confused on how to solve this problem since you will have to impliciltly differntiate the equation and idk what comes after to find the points. The answer to this is (-3,1) pls show solution and explain

 

Answer 1

The curve y³ + x² = y² - 5y + 14

Differentiate implicitly with respect of x.

3y² y' + 2x = 2yy' - 5y'

3y² y' + 2x - 2yy' + 5y' = 0

y' (3y² - 2y + 5) + 2x = 0

y' = - 2x/[3y² - 2y + 5]

y' is slope of tangent line

y' = 1

1 = - 2x/[3y² - 2y + 5]

3y² - 2y + 5 = -2x

x = [3y² - 2y + 5]/-2

x = - 3y²/2 + y - 5/2

Substitute the x = - 3y²/2 + y - 5/2  in the curve.

Solving above fourth degree equation, the roots are

y =  0.36 +  1.98i

y = 0.36 - 1.98i

y = 0.99

y = -0.84

In this case consider y = 0.99 = 1(approximate)

Substitute y = 1 in x = - 3y²/2 + y - 5/2

Therefore, the tangency point is (x,y) = (-3, 1).