Use implicit differentiation to find the points where the circle defined by x2+y2−2x−4y=−1 has horizontal tangent lines. List your answers as points in the form (a,b). 1. Find the points where the curve has a horizontal tangent line.How do I solve this question? I know i have to solve for y' to find the gradient of the slope which I calculate to be y' = (2x-2)/(2y-4) y' = x-1/y-2What do I do after this?

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Use implicit differentiation to find the points where the circle defined by x2+y2−2x−4y=−1 has horizontal tangent lines. List your answers as points in the form (a,b). 1. Find the points where the curve has a horizontal tangent line.How do I solve this question? I know i have to solve for y&#039; to find the gradient of the slope which I calculate to be y&#039; = (2x-2)/(2y-4) y&#039; = x-1/y-2What do I do after this?

agyalapi60 · Accepted Answer

Your circle   γ is a level line of the function  F  (  x  ,  y  )  :=      x    2    +      y    2    −  2  x  −  4  y     .Therefore at any point   P  =  (  x  ,  y  )  ∈  γ the gradient   ∇  F  (  P  )  =  (  2  x  −  2  ,  2  y  −  4  ) is orthogonal to the tangent at   P. Since we want the points   P  ∈  γ where the tangent is horizontal we want the points where the normal   ∇  F  (  P  ) is vertical, i.e., where   2  x  −  2  =  0, or   x  =  1. Putting   x  =  1 in the equation for   γ leads to the equation  1  +      y    2    −  2  −  4  y  =  −  1with the two solutions       y    1    =  0 and       y    2    =  4. It follows that there are two points where   γ has a horizontal tangent, namely the points       P    1    =  (  1  ,  0  ) and       P    2    =  (  1  ,  4  ).

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