Question

The angle between curves $y^2=4x$ and $x^2+y^2=5$ at (1, 2) is
[Karnataka CET 1999]

• A. $ta{n}^{-1}\left(3\right)$
• B. $ta{n}^{-1}\left(2\right)$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A $ta{n}^{-1}\left(3\right)$

$y=ta{n}^{-1}\left[\frac{x}{1+\sqrt{1-x^2}}\right]$
$Putx=sin\theta \Rightarrow y=ta{n}^{-1}\left[\frac{sin\theta }{1+cos\theta }\right]=ta{n}^{-1}tan\frac{\theta }{2}=\frac{\theta }{2}$
$\Rightarrow y=\frac{1}{2}si{n}^{-1}xandletz=si{n}^{-1}x$
Hence $\frac{dy}{dz}=\frac{\left(\frac{dy}{dx}\right)}{\left(\frac{dz}{dx}\right)}=\frac{\frac{1}{2}\frac{d}{dx}si{n}^{-1}x}{\frac{d}{dx}si{n}^{-1}x}=\frac{1}{2}.$