in the equation x^2+xy+y^2=7 the curve has two x-intercepts. Find the equations

Question

in the equation x^2+xy+y^2=7 the curve has two x-intercepts. Find the equations of the lines tangent to the curve at those two x-intercepts. Please show all steps. Thanks

Explanation / Answer

By letting y = 0, we see that x^2 + xy + y^2 = 7 intercepts the x-axis when: x^2 + 0 + 0 = 7 ==> x = ±v7. So, your two x-intercepts are (-v7, 0) and (v7, 0). Then, by differentiating implicitly: (d/dx)(x^2 + xy + y^2) = (d/dx)(7) ==> 2x + x(dy/dx) + y + 2y(dy/dx) = 0, by the Product and Chain rules ==> (x + 2y)(dy/dx) = -2x - y, by bringing all non-dy/dx terms to one side ==> dy/dx = (-2x - y)/(x + 2y), by solving for dy/dx. If y = 0, then dy/dx = -2. Thus, the slope of the tangent line at both points is -2. Using point-slope form, the two tangent lines are: (a) At (-v7, 0): y - 0 = -2[x - (-v7)] ==> y = -2x - 2v7 (a) At (v7, 0): y - 0 = -2(x - v7) ==> y = -2x + 2v7.

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