Question

Find the two points where the curve \(x^{2}+x y+y^{2}=7\) crosses the \(x\) -axis, and show that the tangents to the curve at these points are parallel. What is the common slope of these tangents?

Short answer

The two points where the curve crosses the x-axis are (-sqrt(7), 0) and (sqrt(7), 0). The common slope of the tangents at these points is \( \frac{-1}{2\sqrt{7}} \). Thus, the tangents are parallel.

Step-by-step solution

Find the x-intercepts

Set \(y=0\) in the equation \(x^{2}+x y+y^{2}=7\), the equation simplifies to \(x^{2}=7\). Solving this equation we get \(x=\pm \sqrt{7}\). Hence, our points are \((- \sqrt{7},0)\) and \((\sqrt{7},0)\)

Find the slope of the tangents

The slope of the tangent at any point (x, y) on the curve is given by the negative reciprocal of dx/dy. To find dx/dy, we differentiate both sides of the equation \(x^{2}+x y+y^{2}=7\) with respect to y, while treating x as a function of y. The derivative is \(2x \frac{dx}{dy}+x+y+2y \frac{dx}{dy}=0\). Since we are interested in the points where y=0, we substitute \(y = 0\) and \(x = \pm \sqrt{7}\) into the equation, which then simplifies to \(2\dfrac{dx}{dy}=\dfrac{-1}{\sqrt{7}}\). Hence the slope of the tangent at both points is \( \frac{-1}{2\sqrt{7}} \)

Check if the tangents are parallel

Two lines are parallel if and only if their slopes are equal. The slope of the tangent at both points is \( \frac{-1}{2\sqrt{7}} \), hence the tangents are indeed parallel