Question

A tangent line is drawn to the hyperbola $xy=c$ at a point $P$ , how do you show that the midpoint of the line segment cut from this tangent line by the coordinate axes is $P$ ?

Answer 1

In this question we have been asked to prove that the midpoint of the tangent of the hyperbola $xy=c$ is at $P$ where it touches the hyperbola. We will derive the slope of tangent which is given by differentiating the hyperbola equation and substituting the coordinates of the point. The equation of the line passing through $\left( {{x}_{1}},{{y}_{1}} \right)$ having slope $m$ is given as $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ . The midpoint of a line segment passing through $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ will be given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ .

Complete step-by-step solution:
Now considering from the question we have to show that the midpoint of the line segment cut from this tangent line by the coordinate axes is $P$ . This is a tangent line drawn to the hyperbola $xy=c$ at a point $P$ .
Now we will differentiate the given equation of the hyperbola $xy=c$ then we will have $\Rightarrow x\left( \dfrac{dy}{dx} \right)+y=0\Rightarrow \dfrac{dy}{dx}=\dfrac{-y}{x}$ .
Let us assume that the $x$ coordinate of $P$ is $t$. So the $y$ coordinate of $P$ will be given as $\dfrac{c}{t}$ . So $P$ is $\left( t,\dfrac{c}{t} \right)$ .
So the slope of the tangent at $P$ will be given as $\Rightarrow {{\dfrac{dy}{dx}}_{x=t}}=\dfrac{\left( \dfrac{-c}{t} \right)}{t}\Rightarrow \dfrac{-c}{{{t}^{2}}}$ .
From the basic concepts we know that the equation of the line passing through $\left( {{x}_{1}},{{y}_{1}} \right)$ having slope $m$ is given as $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ .
Hence the equation of the tangent at $P$ will be $\left( y-\dfrac{c}{t} \right)=\left( \dfrac{-c}{{{t}^{2}}} \right)\left( x-t \right)$ .
We can simplify this and write it as
$\begin{aligned} & \Rightarrow \left( \dfrac{ty-c}{t} \right)=\left( \dfrac{-c}{{{t}^{2}}} \right)\left( x-t \right) \\\ & \Rightarrow \left( ty-c \right)=\left( \dfrac{-c}{t} \right)\left( x-t \right) \\\ & \Rightarrow t\left( ty-c \right)=-c\left( x-t \right) \\\ & \Rightarrow {{t}^{2}}y-ct=-cx+tc \\\ & \Rightarrow cx+{{t}^{2}}y-2ct=0 \\\ \end{aligned}$
So the equation of the tangent at $P$ is $cx+{{t}^{2}}y=2ct$ .
When $x=0$ the tangent will cut the $x$ axis at
$\begin{aligned} & c\left( 0 \right)+{{t}^{2}}y=2ct\Rightarrow {{t}^{2}}y=2ct \\\ & \Rightarrow y=\dfrac{2ct}{{{t}^{2}}}\Rightarrow y=\dfrac{2c}{t} \\\ \end{aligned}$ .
When $y=0$ the tangent will cut the $y$ axis at $cx+{{t}^{2}}\left( 0 \right)=2ct\Rightarrow x=2t$ .
So the tangent touches the axis at $\left( 2t,0 \right)$ and $\left( 0,\dfrac{2c}{t} \right)$ .
From the basics concept the midpoint of a line segment passing through $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ will be given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ .
So the midpoint will be located at $\Rightarrow \left( \dfrac{2t+0}{2},\dfrac{\left( \dfrac{2c}{t}+0 \right)}{2} \right)=\left( t,\dfrac{c}{t} \right)$ .
Hence it is proved that the midpoint of the tangent of the hyperbola $xy=c$ is located at $P\left( t,\dfrac{c}{t} \right)$ .

Note: During solving this type of questions we should be sure with our concepts and calculations that we perform while solving questions of this type. The general equation of a line passing through $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $\left( y-{{y}_{1}} \right)=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)$ where the slope is given as $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .