Answer
$(\pm\sqrt 7,0)$,
Slope is $-2$
Work Step by Step
Step 1. To find the two points where the curve $x^2+xy+y^2=7$ crosses the x-axis, let $y=0$; we have $x^2=7$, which gives $x=\pm\sqrt 7$ and the two points are $(\pm\sqrt 7,0)$
Step 2. To find the slopes of the tangents to the curve, differentiate both sides of the equation with respect to $x$:
$2x+y+xy'+2yy'=0$, which gives $y'=-\frac{2x+y}{x+2y}$
Step 3. For point $(\sqrt 7,0)$, slope $m_1=-\frac{2\sqrt 7+0}{\sqrt 7+0}=-2$
Step 4. For point $(-\sqrt 7,0)$, slope $m_1=-\frac{-2\sqrt 7+0}{-\sqrt 7+0}=-2$
Step 5. Thus the two tangent lines are parallel with a common slope of $-2$
