Find 2 coordinates of the points where the function has a horizontal tangent line? 2y^2+4x^2-y=16x

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Feb 9, 2018

Answer:

Tangent is horizontal at #(2,3.09)# and #(2,-2.59)#

Explanation:

The function #2y^2+4x^2-y=16x# will have a horizontal tangent line, where slope of the curve i.e. derivative of function equals zero.

Hence let us first find the derivative of #2y^2+4x^2-y=16x#. Differentiating (considering it an implicit function), we get

#4y(dy)/(dx)+8x-(dy)/(dx)=16#

or #(dy)/(dx)(4y-1)=16-8x#

or #(dy)/(dx)=(16-8x)/(4y-1)=(8(2-x))/(4y-1)#

and this is zero when #x=2#.

When #x=2#, we have #2y^2+4*2^2-y=16xx2#

or #2y^2-y+16-32=0#

or #2y^2-y-16=0#

i.e. #y=(1+-sqrt(1+128))/4=(1+-sqrt129)/4#

i.e. #3.09# or #-2.59#

Hence tangent is horizontal at #(2,3.09)# and #(2,-2.59)#

graph{(2y^2+4x^2-y-16x)((x-2)^2+(y-3.09)^2-0.02)((x-2)^2+(y+2.59)^2-0.02)=0 [-10, 10, -5, 5]}

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