Question

The acute angles between the curves $y=2x^{2}-x$ and $\text{y}^{\text{2}}=\text{x}$ at $\left(0,0\right)$ and $\left(1,1\right)$ are $\alpha $ and $\beta $ respectively, then

• A. $\alpha -\beta =0$
• B. $\alpha +\beta =0$
• C. $\alpha >\beta $
• D. $\alpha < \beta $

Answer 1

Answer: (A)

$y=2x^{2}-x$
$\Rightarrow \frac{d y}{d x}=4x-1$
Let, $m_{1}=4x-1$
$y^{2}=x$
$\Rightarrow 2y\frac{d y}{d x}=1$
Let, $m_{2}=\frac{d y}{d x}=\frac{1}{2 y}$
At $P\left(0,0\right),m_{1}=-1,m_{2} \rightarrow $ not defined
Angle $\alpha =\frac{\pi }{4}$
At $Q\left(1,1\right),m_{1}=3,m_{2}=\frac{1}{2}$
$\beta =\left(t a n\right)^{- 1}\left(\frac{3 - \frac{1}{2}}{1 + \frac{3}{2}}\right)=\frac{\pi }{4}$