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Question

Find the equation of the tangent and the normal to the following curve at the indicated point.
$$xy=c^2$$ at $$\left (ct, \dfrac {c}{t}\right)$$.


Solution

$$xy=c^2$$
Differentiating both sides w.r.t. $$x,$$
$$\Rightarrow$$  $$x\dfrac{dy}{dx}+y=0$$
$$\Rightarrow$$  $$\dfrac{dy}{dx}=\dfrac{-y}{x}$$
$$\Rightarrow$$  $$(x_1,y_1)=\left(ct,\dfrac{c}{t}\right)$$
Slope of tangent, $$m=\left(\dfrac{dy}{dx}\right)_{\left(ct,\dfrac{c}{t}\right)}=\dfrac{\dfrac{-c}{t}}{ct}=\dfrac{-1}{t^2}$$
Equation of tangent is,
$$y-y_1=m(x-x_1)$$
$$\Rightarrow$$  $$y-\dfrac{c}{t}=\dfrac{-1}{t^2}(x-ct)$$

$$\Rightarrow$$  $$\dfrac{yt-c}{t}=\dfrac{-1}{t^2}(x-ct)$$

$$\Rightarrow$$  $$yt^2-ct=-x+ct$$
$$\Rightarrow$$  $$x+yt^2=2ct$$
Equation of normal is,
$$y-y_1=\dfrac{-1}{m}(x-x_1)$$
$$\Rightarrow$$  $$y-\dfrac{c}{t}=t^2(x-ct)$$
$$\Rightarrow$$  $$yt-c=t^3x-ct^4$$
$$\Rightarrow$$  $$xt^3-yt=ct^4-c$$
$$\therefore$$  Equation of tangent is $$x+yt^2=2ct$$.
$$\therefore$$  Equation of normal is $$xt^3-yt=ct^4-c$$

Mathematics

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