Slope of Tangent line?

Question

find the slop of a tangent line at the indicated point. write an equation fot the tangent line.

x^2-y^2=1 at (3,1)

Answer 1

y^2=x^2+1
2y(dy/dx)=2x
dy/dx=x/y=3/1
y=3x+b
1=3(3)+b
b=-8
y=3x-8

Answer 2

Beware of people who calculate anwers blindly without checking the facts. First note that (3,1) is not on the graph. Check by putting 3 in for x and solving for y to get +/-sqrt(8), so you can't find the tanget at (3,1). However you can find the slope of the tanget at (3,sqrt(8)) or [3, 2sqrt(2)].

Differentiate with respect to x to get:

2x - 2y dy/dx = 0 and solving for slope = dy/dx = x/y

So the slope at (3,sqrt(8)) is

m = dy/dx = x/y = 3/sqrt(8)

You can simplify this answer to 3sqrt(2) / 4

Now use y - y1 = m ( x - x1 ) to find the equation of the tangent line.

Answer 3

2x -2yy'=0
y' = y/x = 1/3 at (3,1)
y= x/3 + b
1= 1+b --> b =0
y= x/3 is equation of tangent line.

Answer 4

2x - 2y (dy/dx) = 0
2y (dy/dx) = 2 x
dy/dx = x/y
dy/dx = 3 / 1
dy/dx = 3 (slope)
y - 1 = 3 (x - 3)
y = 3x - 8 (equation of line)