How do you find the points where the graph of the function #x^2+7y^2-4x-2=0# has horizontal tangents?

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Answer 1 · 2016-03-26T12:44:42

At #(2,0.926)# and #(2,-0.926)#, graph of the function
#x^2+7y^2-4x-2=0# will have horizontal tangents.

The graph of the function #x^2+7y^2-4x-2=0# will have horizontal tangents, where its slope is zero i.e. first derivative is equal to zero.

For this, factorizing #x^2+7y^2-4x-2=0# gives us

#2x+14y(dy)/(dx)-4=0# or #(dy)/(dx)=(4-2x)/(14y)#, which will be zero for

#(4-2x)/(14y)=0# or #4-2x=0# if #y!=0# i.e. #x=2#.

Now putting #x=2# in #x^2+7y^2-4x-2=0# we get

#2^2+7y^2-4xx2-2=0# or #4+7y^2-8-2=0#

or #7y^2=10-4=6# or #y=sqrt(6/7)=+-0.926#

Hence points are #(2,0.926)# and #(2,-0.926)#

graph{x^2+7y^2-4x-2=0 [-2, 5, -2, 2]}