How do you find the slope of the line tangent to #x^2-y^2=3# at (2,1), (2,-1), (sqrt3,0)?

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Answer 1 · 2017-05-07T18:09:24

Differentiate each term with respect to x.
Solve for #dy/dx# as a function of #(x,y)#
Evaluate #m = dy(x,y)/dx at the points.

Given: #x^2-y^2=3#

Differentiate each term with respect to x:

#(d(x^2))/dx - (d(y^2))/dx= (d(3))/dx#

#2x - 2ydy/dx = 0#

Solve for #dy/dx#

#-2ydy/dx = -2x#

#dy/dx = x/y#

At the point #(2,1)#, the slope of the tangent line is:

#m = 2/1#

#m = 2#

At the point #(2,-1)#, the slope of the tangent line is:

#m = 2/-1#

#m = -2#

At the point #(sqrt3,0)#, the slope of the tangent line is a division by 0 situation, which indicates a vertical line.