A curve has the equation 2x^2 + xy - y^2 +18 = 0. (1) Find the coordinates of the points where the tangent to the curve is parallel to the x-axis.

Firstly notice that the coordinates of the points where the tangent to the curve is parallel to the x-axis are precisely the points where the rate of change of y with respect to x is not changing. That is, when dy/dx = 0. 

So, differentiating euqation (1) with respect to x gives:

4x + y + xdy/dx - 2ydy/dx = 0, where underline denotes the use of implicit differentiation.

Now, dy/dx = 0 gives us:

4x + y + 0, i.e. y = -4x. 

Substituting this back into euqation (1) and solving the quatratic for x gives x = +/- 1. Thus y = -/+ 4.

Hence, coordinates of the points where the tangent to the curve is parallel to the x-axis are (1,-4), (-1,4).

Answered by Oliver L. Maths tutor

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