Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Equation of langent to the curve $x = \cos\theta, \;y=\sin\theta$ at $\theta = \frac{\pi}{}4$ is

Option 1)
$3x-y =\sqrt{2}$

Option 2)
$3x+y =2\sqrt{2}$

Option 3)
$x -y = 0$

Option 4)
$x+y =\sqrt{2}$

Answers (1)

best_answer

As we have learned

Equation of Tangent for parametric form -

$(y-f_{y}(t))=\frac{f'_{y}}{f'_{x}}(x-f_{x}(t))$

-

For $\theta = \pi /4;$ point will be $\left ( 1/\sqrt 2 ,1/\sqrt2\right )$ , we need slope using $\frac{dy}{dx}$

$\therefore$ slope at $\theta = \pi /4=\frac{dy}{dx}$ at $\theta = \pi /4=\frac{dy/d\theta }{dx/d\theta }$ at $\theta = \pi /4=\frac{\cos \theta }{-\sin \theta }$ at

$\theta = \pi /4$ $= -1$

$\therefore slope =-1$ and point is $\left ( 1/\sqrt 2 ,1/\sqrt2\right )$

$\therefore equation is: (y-1/2)=-1( x-1/\sqrt2)$

$\Rightarrow x+y= \sqrt 2$

Option 1)

$3x-y =\sqrt{2}$

Option 2)

$3x+y =2\sqrt{2}$

Option 3)

$x -y = 0$

Option 4)

$x+y =\sqrt{2}$

Posted by

Himanshu

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in

anam-khan

JEE Mains 2026 LIVE: NTA reopens session 2 registrations; city slip soon; admit card, exam dates

anam-khan

JEE Main 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow

Latest Question