x^2+2xy-y^2+11x=9 find equation of tangent line to curve at point (1,3)?

Answer 1

First check to see if the point is on the curve.

x² + 2xy - y² + 11x = 9
1² + 2*1*3 - 3² + 11*1 = 1 + 6 - 9 + 11 = 9

So the point is on the curve.

Take the derivative of the curve to find the slope. Differentiate implicitly.

2x + 2y + 2x(dy/dx) - 2y(dy/dx) + 11 = 0
(2x - 2y)(dy/dx) = -(2x + 2y + 11)

dy/dx = -(2x + 2y + 11) / (2x - 2y)

Plug in the values.

dy/dx = -(2*1 + 2*3 + 11) / (2*1 - 2*3)
dy/dx = (2 + 6 + 11) / (2 - 6) = -19/4

The slope of the tangent line is the same as the slope of the curve at the point of tangency. The equation of the tangent line is:

y - 3 = (-19/4)(x - 1)
y - 3 = (-19/4)x + 19/4
y = (-19/4)x + 31/4

Answer 2

y=mx+b
m=slope of the line
b=y-intercept