Can we find the triangle areas given by a tangent?

Can we find the triangle areas given by a tangent?

Question

Find the equation of the tangent to the rectangular hyperbola $xy=c^2$ at the point $P(ct,c/t)$.

The tangent at $P$ meets the axes of $x$ and $y$ at $L$ and $M$ respectively. $O$ is the centre of the hyperbola and $POQ$ is a diameter. The straight line $MQ$ meets the axis of $x$ in $T$. Prove that

1. the area of triangle $MOL$ is $2c^2$,

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1. the area of triangle $QOT$ is $c^2/3$.

Suggestion

A well drawn diagram will help us a lot in this question!

What is the gradient at any point on the curve $xy=c^2$?

Can we use the equation of the tangent at $P$ to find $L$ and $M$?

What are the coordinates of $Q$ if $POQ$ is a diameter?

Note that a diameter of a hyperbola is any straight line passing through its centre. We can assume here that $Q$ is the point on the hyperbola opposite $P$. Note; $AOB$ is a diameter of a hyperbola if and only if the straight line segment $AOB$ has rotational symmetry of order $2$ about $O$.

Solution

Differentiating the curve $y=\dfrac{c^2}{x}$ we find

$$\dfrac{dy}{dx}=-\dfrac{c^2}{x^2}.$$

So the gradient at $P = (ct,c/t)$ is $-\dfrac{1}{t^2}$.

Thus the equation of the tangent here is

$$\begin{align*} y-\dfrac{c}{t} &= -\dfrac{1}{t^2}(x-ct) \\ \Longrightarrow y &= -\dfrac{x}{t^2}+\frac{2c}{t} = \dfrac{2ct-x}{t^2} \end{align*}$$

1. Prove that the area of the triangle $MOL$ is $2c^2$

The $x$-intercept of the tangent: $0=\dfrac{2ct-x}{t^2}\quad \Longrightarrow \quad x=2ct,$ so $L$ is $(2ct,0)$.

The $y$-intercept of the tangent: $y=\dfrac{2ct}{t^2}=\dfrac{2c}{t},$ so $M$ is $(0,\dfrac{2c}{t}).$

So using area = $\dfrac{1}{2} \times \,\text{base}\, \times\, \text{height}$, the area of triangle $MOL$ is $\dfrac{1}{2}\times 2ct \times \dfrac{2c}{t}=2c^2.$

1. Prove that the area of the triangle $QOT$ is $c^2/3$.

The coordinates of point $Q$ are $(-ct,-c/t)$ since it is the reflection of $P$ in $O$.

The line joining $Q = (-ct,-c/t)$ and $M = (0,2c/t)$ has gradient $\dfrac{\dfrac{2c}{t}+\dfrac{c}{t}}{ct}=\dfrac{3}{t^2}.$ Therefore the equation of the line $QM$ is $y-\dfrac{2c}{t}={3}{t^2}x.$ The $x$-intercept is then $x=-\dfrac{2c}{t}\times\dfrac{t^2}{3}=-\dfrac{2ct}{3}.$

Thus triangle $QOT$ has base $\dfrac{2ct}{3}$ and height $\dfrac{c}{t}$ so its area is $\dfrac{1}{2}\times \dfrac{2ct}{3}\times\dfrac{c}{t}=\dfrac{c^2}{3}.$