Tangent line of a hyperbola

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

x2 + 2xy − y2 + x = 39, (5, 9) This is a hyperbola.

I am ot sure what to do with all the extra numbers. I know a hyperbola is x^2/a^2 + y^2/b^2=1 But I don't know how to apply this with y'?

x² + 2xy − y² + x = 39

Differentiating to get the slope of the tangent line:
2x + 2xy' + 2y - 2yy' + 1 = 0
(2x-2y)y' + 2x+2y+1 = 0
y' = -(2x+2y+1)/(2x-2y)

Verify (5,9) is on the curve:
25+2(5*9) - 81 + 5 = 39
25+90+5 - 81 = 39
120-81 = 39
39 = 39
The given point is on the curve.

The slope of the tangent line is:
y' = -(2x+2y+1)/(2x-2y)
y' = -(2*5+2*9+1)/(2*5-2*9) = -(10+18+1)/(10-18) = 29/8

The tangent line (in standard form) is:
y-9 = (29/8)(x-5)
8y-72 = 29x - 145
29x - 8y = 73

The line in point slope form is:
y = (29/8)x - 73/8

The fact that this is a hyperbola is of little importance.

x^2 + 2xy - y^2 + x = 39 --> Implicitly Differentiate.

2x + (2)y + 2x(dy / dx) - 2y(dy /dx) + 1 = 0 --> Subtract 2x from both sides.

2y + 2x(dy / dx) - 2y(dy / dx) + 1 = - 2x --> Subtract 2y from both sides.

2x(dy / dx) - 2y(dy / dx) + 1 = - 2x - 2y --> Subtract 1 from both sides.

2x(dy / dx) - 2y(dy / dx) = - 2x - 2y - 1 --> Factor out a common dy / dx.

(dy / dx) [ 2x - 2y ] = - 2x - 2y - 1 --> Divide through by (2x - 2y).

(dy / dx) = (- 2x - 2y - 1) / (2x - 2y) --> Substitute x = 5 and y = 9 (the point given).

(dy / dx) = [ - 2(5) - 2(9) - 1 ] / [ 2(5) - 2(9) ] --> Simplify.

(dy / dx) = 29 / 8

Note: dy / dx is the slope of the tangent line. Recognize, now that you have the slope, and you are given a point...you can use "Point Slope Formula" to solve for the equation of the tangent line.

y - y1 = m(x - x1) --> Substitute all values in.

y - 9 = (29 / 8)(x - 5) --> Distribute.

y - 9 = (29 / 8)x - (145 / 8) --> Add 9 to both sides.

ANSWER: y = (29 / 8)x - (73 / 8)

Note: Anytime you see a product of xy together, you must perform the Product Rule.