# Find the points where the curve x^2 - xy + y^2 = 3 crosses the x-axis. Show that at these points, the tangent lines to the curve are parallel.

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You should remember that y=0 if the curve crosses x axis, hence, substituting y=0 in the given equation yields:

`x^2 - x*0 + 0^2 = 3 => x^2 = 3 => x_(1,2) = +-sqrt3`

Hence, the curve intercepts x axis atc `(-sqrt3, 0)` and `(sqrt 3,0) .`

You need to prove that the tangent lines to the curve, at these points, are parallel, hence, you should check if derivatives at `x = +-sqrt3` have equal values such that:

`2x - (y + x(dy)/(dx)) + 2y(dy)/(dx) = 0`

`2x - y - x(dy)/(dx) + 2y(dy)/(dx) = 0`

You need to isolate the terms that contain `(dy)/(dx)` to the left side such that:

`-x(dy)/(dx) + 2y(dy)/(dx) = y - 2x`

Factoring out `(dy)/(dx)` yields:

`(dy)/(dx)(-x + 2y) = y - 2x =>(dy)/(dx) = (y-2x)/(2y-x)`

You need to evaluate `(dy)/(dx)` at `(-sqrt3,0)` such that:

`(dy)/(dx)|_(-sqrt3,0) = (0 + 2sqrt3)/(0 + sqrt3) = 2`

`(dy)/(dx)|_(sqrt3,0) =(0-2sqrt3)/(0-sqrt3) = ` 2

**Since the slopes of the tangent lines have the same values, `(dy)/(dx)|_(-sqrt3,0) = (dy)/(dx)|_(sqrt3,0) = 2` , then the tangent lines to the curve, at the points `(-sqrt3, 0)` and `(sqrt 3,0)` are parallel**.