If -xy+x=-5y^2-5 then find the equations of all tangent lines to the curve when x=-5.

Question

iimplicit derivative at a point

Answer 1

When x = -5:

-(-5)y + (-5) = -5y²-5

5y² + 5y = 0

y(y+1) = 0

y=0 or y = -1

The following points are the points of tangency: (-5,0) and (-5,-1).

Finding the derivative implicitly:

-xy' + y(-1) + 1 = -10yy'

10yy' -xy' = y - 1

y'(10y - x) = y-1

y' = (y-1)/(10y - x)

To find the slopes of the tangent lines:

y'(-5,0) = -1/-(-5) = -1/5

y'(-5,-1) = -2/-5 = 2/5

Using point slope for equations of tangent lines at those points:

(-5,0):

y - 0 = -1/5(x + 5)

y = -1/5 x - 1

(-5,-1):

y + 1 = 2/5(x + 5)

y = 2/5 x + 1

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