At what x value does the curve #x^2+ 7y^2- 4x- 2 = 0# have a horizontal tangent?

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Answer 1 · 2015-10-12T13:51:23

See the explanation section below. (I made my best guess for the intended question and change the #+0# to #=0#)

No Calculus
#x^2+ 7y^2- 4x- 2 = 0# is the equation of an ellipse.
The tangent will be horizontal at the #x# coordinate of the center.

Completing the square gets us:#(x-2)^2+ 7y^2 = 6#

The tangent lines will be horizontal at #x=2#

Implicit differentiation
#x^2+ 7y^2- 4x- 2 = 0#

#2x + 14y dy/dx -4 = 0#

#dy/dx = (4-2x)/(14y)#

#dy/dx = 0# at #x=2#