Question

The angle of intersection of the curves $y=x^2$ and $x=y^2$ at (1, 1) is
[Roorkee 2000; Karnataka CET 2001]

• A. $ta{n}^{-1}\left(\frac{4}{3}\right)$
• B. $ta{n}^{-1}\left(1\right)$
• C. ${90}^o$
• D. $ta{n}^{-1}\left(\frac{3}{4}\right)$

Answer 1

The correct option is C ${90}^o$

$x=logp\Rightarrow p=e^x\Rightarrow y=e^{-x}$
$\Rightarrow \frac{dy}{dx}=e^{-x}and\frac{d^{2}y}{d{x}^2}=e^{-x};\therefore \frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=0.$