How to prove the triangle is isosceles and determine its area

Question

Let $P(a,b)$ be a point on the curve $y = \frac{1}{x}$ in first quadrant and let the tangent line at $P$ intersect the $x$-axis at a point $A$. Show that triangle $AOP$ is isosceles and determine its area, where $O$ is the origin.

How do I start to attempt this?

Answer 1

To start approaching the problem, first notice that if P lies on the curve then $P (a, b) = P (a, \frac{1}{a})$.

Secondly, find the slope of the tangent line at $P$, which is precisely $-\frac{1}{a^2}$

Thus equation of the tangent line through $P$ is given by

$$(y - \frac{1}{a}) = (-\frac{1}{a^2})(x-a) \implies a^2y = 2a - x$$

The above line intersects the $x$-axis only at $(2a, 0)$, thus $\ A = (2a, 0)$.

Therefore the coordinates of $\Delta AOP$ are precisely: $$A (2a, 0), \ \ O(0, 0), \ \ P (a, \frac{1}{a})$$ making $AOP$ isosceles. Moreover, the area of $\Delta AOP = \frac{1}{2} \cdot 2a \cdot \frac{1}{a} = 1$ unit.

Answer 2

Hint:

WLOG $b=1/a>0$

$$xy-1\implies x\dfrac{dy}{dx}+y=0\implies\dfrac{dy}{dx}_{\text{ at }(a,b)}=-\dfrac{1/a}a=-\dfrac1{a^2}$$

The equation of the tangent at $P,$ $$\dfrac{y-1/a}{x-a}=-\dfrac1{a^2}\iff x+a^2y=2a$$

As $x$ axis, $y=0,$ the intersection will be $A(2a,0)$

$$|AP|=?, |OP|=?$$

Answer 3

The tangent line to $y=\frac 1x$ at $P(x_0,y_0)$ is: $$y=y_0+y'(x_0)(x-x_0)=\frac 1{x_0}-\frac 1{x_0^2}(x-x_0).$$ The point $A(x_1,0)$ has coordinate $x_1$: $$0=\frac 1{x_0}-\frac 1{x_0^2}(x_1-x_0) \Rightarrow x_1=2x_0,$$ which shows the triangle $PAO$ is isosceles, because the altitude $PB$ is the median too.

Now the area of $\Delta PAO$ is: $$S=\frac 12 \cdot AO\cdot PB=\frac12 \cdot 2x_0 \cdot \frac1{x_0}=1.$$