Question

Find lenght of subnormal to $x=\sqrt{2}\cos t,y=-3\sin t$ at $t=\frac{-\pi }{4}$

Answer 1

$x=\sqrt{2}cost,y=-3sintatt=\frac{-\pi }{4}\Rightarrow \frac{dx}{dt}=-\sqrt{2}sint,\frac{dy}{dt}=-3cost\frac{dx}{dt}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}=\frac{-3cost}{-\sqrt{2}sint}=\frac{3}{\sqrt{2}}cot\left(t\right)\therefore slope=\frac{3}{\sqrt{2}}cot\left(t\right)$

length of subtangent = $\left|{}_{\left(slpoe\right)(y-coorinate)}\right|$

$att=-\frac{\pi }{4},=y-3sin(-\frac{\Pi }{4})=3\times -\frac{1}{\sqrt{2}}=\frac{3}{\sqrt{2}}$

$\therefore$ length of subtangent = $\left|\frac{3}{\sqrt{2}}cot\right(t)\cdot \frac{3}{\sqrt{2}}|=\left|\frac{9}{2}cot\left(\frac{-\pi }{4}\right)\right|=\frac{9}{2}$