x^2 - 3xy + y^2 = 1
2x -3x(dy/dx) - 3y + 2y dy/dx = 0
dy/dx(-3x + 2y) = 3y - 2x
dy/dx = (3y - 2x)/(2y - 3x)
to have a horizontal slope, dy/dx = 0 , that is,
3y - 2x = 0
y = 2x/3
sub that back into the original
x^2 - 3x(2x/3) + 4x^2/9 = 1
9x^2 - 18x^2 + 4x^2 = 1
-5x^2 = 1
There is no real solution to this, so there is no such tangent.
find the derivative like I showed you in the first two.
Since the tangent is vertical, the denominator of the slope must be zero.
Show that this is not possible.
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