Question

$x=$ $\frac{1}{t}$ , $y=t-$$\frac{1}{t}$ at $t=2$.Find tangent and normal line.

Answer 1

$x=\frac{1}{t}\frac{dx}{dt}=\frac{-1}{t^2}$

$\frac{dx}{dt}=\frac{-1}{4}(t=2)y=t-\frac{1}{t}\frac{dy}{dt}=1+\frac{1}{t^2}\frac{dy}{dt}=1+\frac{1}{4}=\frac{5}{4}(t=2)\therefore \frac{dx}{dy}=\frac{\frac{5}{4}}{\frac{-1}{4}}=-5x=\frac{1}{2}y=2-\frac{1}{2}=\frac{3}{2}$

slope of tangent $=-5$

slope of normal $=\frac{1}{5}$

equation of tangent

$y-\frac{3}{2}=-5(x-\frac{1}{2})5x+y=4$

equation of normal

$y-\frac{3}{2}=\frac{1}{5}(x-\frac{1}{2})x-5y+7=0$