Question

The acute angle between curves $x^2+y^2=4$ and $x^2=3y$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. ${\tan }^{-1}\left(2\right)$
• D. ${\tan }^{-1}\left(\frac{5}{\sqrt{3}}\right)$

Answer 1

The correct option is D ${\tan }^{-1}\left(\frac{5}{\sqrt{3}}\right)$

$y^2+3y=4$ (from the $e{q}^n$)
$y^2+3y-4=0$
$y^2+4y-y-4=0$
$(y^2+4y)-(y-1)=0$
$y-1,x=\sqrt{3},-\sqrt{3}$
$y=-4$
$y^1=\frac{x}{y}=m_1=-\sqrt{3},\sqrt{3}$
$y^1=\frac{2x}{3}=m_2=2\sqrt{3},-2\sqrt{3}$
$tan\theta =\left(\frac{m_2-m_1}{1+m_1{m}_2}\right)$
$=\left(\frac{2\sqrt{3}+\sqrt{3}}{1-6}\right)$
$=-\left(\frac{3\sqrt{3}}{5}\right)$
$tan\theta =\left(\frac{\sqrt{3}+2\sqrt{3}}{1-6}\right)$
$tan\theta =-\frac{3\sqrt{3}}{5}$