Answer to Question #312320 in Calculus for shh

Question

Find the equations of the tangents to the graph of y=x+1/x that are parallel to y+2x=0

Answer 1

Slope of the line y+2x=0: m=-2.

Slope of the tangent: $y'=1-\frac{1}{x^2}.$

We have: $y'=m$ or $1-\frac{1}{x^2}=-2.$

So, $x=\frac{1}{\sqrt{3}},\;x=-\frac{1}{\sqrt{3}}.$

$y(\frac{1}{\sqrt{3}})=\frac{1}{\sqrt{3}}+\sqrt{3}.$

$y(-\frac{1}{\sqrt{3}})=-\frac{1}{\sqrt{3}}-\sqrt{3}.$

Equations of the tangent lines:

$y-\frac{1}{\sqrt{3}}-\sqrt{3}=-2(x-\frac{1}{\sqrt{3}})$ or $y=-2x+2\sqrt{3}$ .

$y+\frac{1}{\sqrt{3}}+\sqrt{3}=-2(x+\frac{1}{\sqrt{3}})$ or $y=-2x-2\sqrt{3}.$