Question

If $x=t^2$and y = 2t, then equation of the normal at t = 1 is
[RPET 1996]

• A. x + y - 3 = 0
• B. x + y - 1 = 0
• C. x + y + 1 = 0
• D. x + y + 3 = 0

Answer 1

The correct option is A x + y - 3 = 0

Let y $=ta{n}^{-1}\sqrt{\frac{1-x^2}{1+x^2}}andz=co{s}^{-1}\left(x^2\right)$
Put $x^2=cos2\theta ;$ $\therefore y=ta{n}^{-1}\sqrt{\frac{2si{n}^2\theta }{2co{s}^2\theta }}=\theta andz=2\theta$
$\therefore \frac{dy}{dz}=\frac{dy/d\theta }{dz/d\theta }=\frac{1}{2}.$