Question

Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x > 0)

Answer 1

Answer:

Step-by-step explanation:

given a point $(x_0,y_0)$ the equation of a line with slope m that passes through the  given point is

$y-y_0 = m(x-x_0)$ or equivalently

$y = mx+(y_0-mx_0)$.

Recall that a line of the form $y=mx+b$, the y intercept is b and the x intercept is $\frac{-b}{m}$.

So, in our case, the y intercept is $(y_0-mx_0)$ and the x  intercept is $\frac{mx_0-y_0}{m}$.

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph $(x_0,\frac{1}{x_0})$. Which means that $y_0=\frac{1}{x_0}$

The slope of the tangent line is given by the derivative of the function evaluated at $x_0$. Using the properties of derivatives, we get

$y' = \frac{-1}{x^2}$. So evaluated at $x_0$ we get $m = \frac{-1}{x_0^2}$

Replacing the values in our previous findings we get that the y intercept is

$(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}$

The x intercept is

$\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0$

The triangle in consideration has height $\frac{2}{x_0}$ and base $2x_0$. So the area is

$\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2$

So regardless of the point we take on the graph, the area of the triangle is always 2.