Question

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

Answer 1

$xy=c^2$

Differentiating both sides w.r.t. $x,$

$\Rightarrow$ $x\frac{dy}{dx}+y=0$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-y}{x}$

$\Rightarrow$ $(x_1,y_1)=(ct,\frac{c}{t})$

Slope of tangent, $m={\left(\frac{dy}{dx}\right)}_{(ct,\frac{c}{t})}=\frac{\frac{-c}{t}}{ct}=\frac{-1}{t^2}$

Equation of tangent is,

$y-y_1=m(x-x_1)$

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

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

$\Rightarrow$ $y{t}^2-ct=-x+ct$

$\Rightarrow$ $x+y{t}^2=2ct$

Equation of normal is,

$y-y_1=\frac{-1}{m}(x-x_1)$

$\Rightarrow$ $y-\frac{c}{t}=t^2(x-ct)$

$\Rightarrow$ $yt-c=t^{3}x-c{t}^4$

$\Rightarrow$ $x{t}^3-yt=c{t}^4-c$

$\therefore$ Equation of tangent is $x+y{t}^2=2ct$.

$\therefore$ Equation of normal is $x{t}^3-yt=c{t}^4-c$