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

Question

William Abdalla · Accepted Answer

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.

Adam Adkins · Answer

undefined To find the slope of the line tangent to the curve x^2 - y^2 = 3 at a given point, we can use the derivative of the equation. Taking the derivative of x^2 - y^2 = 3 with respect to x, we get 2x - 2yy' = 0. Solving for y', we have y' = x/y.

At the point (2,1), the slope of the tangent line is 2/1 = 2.
At the point (2,-1), the slope of the tangent line is 2/-1 = -2.
At the point (√3,0), the slope of the tangent line is (√3)/0, which is undefined.