3

A line is drawn from the origin $O$ to a point $P(x,y)$ in the first quadrant on the graph of $y=\frac{1}{x}$. From point $P$, a line is drawn perpendicular to the $x$-axis, meeting the $x$-axis at $B$. Express the perimeter of $OPB$ as a function of $x$.

I need help setting up the equation for this. Do I just need to determine the equations for each side of the triangle $(a, b, c)$, using the points given? So it would end up looking like $x + y + {}$ hypotenuse. Then wherever I see a $y$, I replace it with $\frac{1}{x}$, giving me $x + \frac{1}{x} + x^2 + (\frac{1}{x})^2$.

Later
  • 722
  • 2
  • 8
  • 24
erimar77
  • 741

1 Answers1

4

You are essentially right. The two "legs" have length $x$ and $1/x$. The hypotenuse is $\sqrt{x^2+1/x^2}$. You just left out the square root symbol. So the perimeter is $$x+\frac{1}{x}+\sqrt{x^2+\frac{1}{x^2}}.$$

André Nicolas
  • 507,029