Review question

# Can we find the triangle areas given by a tangent? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8286

## Solution

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

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 $\dfrac{1}{2}\times \dfrac{2ct}{3}\times\dfrac{c}{t}=\dfrac{c^2}{3}.$