## University Calculus: Early Transcendentals (3rd Edition)

1) The points of intersection between the curve and the $x$-axis are $(\sqrt7,0)$ and $(-\sqrt7,0)$. 2) The tangents to the curve at these points are parallel. They both have the common slope $-2$.
$$x^2+xy+y^2=7$$ 1) Find the derivative of the function using implicit differentiation: $$2x+(y+xy')+2yy'=0$$ $$y'(x+2y)=-(2x+y)$$ $$y'=-\frac{2x+y}{x+2y}$$ 2) Find the points of intersection between the curve and the $x$-axis: - Substitute $y=0$ back to the formula of the function: $$x^2+x\times0+0^2=7$$ $$x^2=7$$ $$x=\pm\sqrt7$$ So the curve crosses the $x$-axis at 2 points $A(\sqrt7,0)$ and $B(-\sqrt7,0)$. 3) Prove that the tangents at $A$ and $B$ are parallel. To prove that the tangents at $A$ and $B$ are parallel, we need to show that the slope of the tangents at these two points are equal. - The slope of the tangent at $A(\sqrt7,0)$ is $$y'=-\frac{2\sqrt7+0}{\sqrt7+2\times0}=-\frac{2\sqrt7}{\sqrt7}=-2$$ - The slope of the tangent at $B(-\sqrt7,0)$ is $$y'=-\frac{2(-\sqrt7)+0}{-\sqrt7+2\times0}=-\Big(\frac{-2\sqrt7}{-\sqrt7}\Big)=-\frac{2\sqrt7}{\sqrt7}=-2$$ Indeed, these tangents have the common slope $-2$, so they are parallel with each other.