Vertical tangent line

For what values does the function 5x^2+2xy+y^2=16 have a vertical tangent line

Expert Answers

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You need to remember that the tangent line to the graph of function is vertical and parallel to y axis if the slope of this tangent line is not defined.

You need to use implicit differentiation to find `(dy)/(dx) ` such that:

`(d(5x^2 + 2xy + y^2))/(dx) = (d(16))/(dx)`

`d/(dx)(5x^2) + d/(dx)(2xy) + d/(dx)(y^2) = 0`

`10x + 2(x*d/(dx)y + y*d/(dx)x) + 2y*(dy)/(dx) =0`

`10x + 2x*(dy)/(dx) + 2y + 2y*(dy)/(dx) = 0`

You need to keep to the left side the terms that contain `(dy)/(dx)`  such that:

`2(dy)/(dx)*(x + y) = -10x - 2y`

`(dy)/(dx)*(x + y) = -5x - y`

`(dy)/(dx) = (-5x - y)/(x+y)`

Hence, the slope `(dy)/(dx) ` is not defined if denominator cancels such that:

x + y = 0 => x = -y

You need to substitute -y for x in equation of function such that:

`5*(-y)^2 - 2y^2 + y^2 = 16 =gt 4y^2 = 16 =gt y^2 = 4`

`y_(1,2) = +-2`

Hence, evaluating the values for the tangent line to the graph of function is vertical yields x = -y and `y = +-2.`

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