Question

Find the equation of the tangent and normal to the given curve at the given points

$x=cost,y=sint=\frac{\pi }{4}$

Answer 1

(v) The equation of the given curve is $x=costy=sint$ ...(i)

The Point on the curve corresponding to $t=\frac{\pi }{4}$ is

$x=cos\frac{\pi }{4},y=sin\frac{\pi }{4}$ $(Onputtingt=\frac{\pi }{4})$

$\Rightarrow x=\frac{1}{\sqrt{2}},y=\frac{1}{\sqrt{2}}$

On differentiating w.r.t. x, we get

$\frac{dx}{dt}=-sint,\frac{dy}{dt}=cost\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{cost}{-sint}=-$cot t

${\left(\frac{dy}{dx}\right)}_{t=\frac{\pi }{4}}=-cot\frac{\pi }{4}=-1$

$\therefore$ Slope of tangent at $t=\frac{\pi }{4}$ is -1.

Hence, the equation of the tangent to the given curve at $t=\frac{\pi }{4}$ i.e.,

at $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})isy-\frac{1}{\sqrt{2}}=-(x-\frac{1}{\sqrt{2}})$

$\Rightarrow x+y-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\Rightarrow x+y-\frac{2}{\sqrt{2}}=0\Rightarrow x+y-\sqrt{2}=0$

A gain, the slope of normal at $t=\frac{\pi }{4}$ is

$\frac{-1}{Slopeoftangentat=\frac{\pi }{4}}=\frac{-1}{-1}=1$

Hence, the equation of the normal to the given curve at $t=\frac{\pi }{4}$

i.e.,at $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})isy-\frac{1}{\sqrt{2}}=1(x-\frac{1}{\sqrt{2}})\Rightarrow x-y=0$.