Question

The angle between curves $x^2+y^2=1$ and $y=4\sqrt{3}{x}^3$ is

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

Answer 1

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

$x^2+y^2=1$ -- (i)
$y=4\sqrt{3}{x}^3$ -- (ii)

On solving (i) and (ii) simultaneously we get,
$x^2+{48x}^6-1=0$
$⟹(4x^2-1)(12x^4+3x^2+1)=0$
$⟹x^2=\frac{1}{4}$
$⟹$ Points of intersection are $(+\frac{1}{2},+\frac{\sqrt{3}}{2})$

Slope of curve $x^2+y^2=1at(\frac{1}{2},\frac{\sqrt{3}}{2})is{m}_1=-\frac{1}{\sqrt{3}}$
Slope of curve $y=3\sqrt{3}{x}^{3}at(\frac{1}{2},\frac{\sqrt{3}}{2})is{m}_2=3\sqrt{3}$
$\Rightarrow$Angle between two curves $\theta ={\tan }^{-1}\left|\frac{-\frac{1}{\sqrt{3}}-3\sqrt{3}}{1+(-3)}\right|$
$={\tan }^{-1}\left|\frac{-10}{-2\sqrt{3}}\right|={\tan }^{-1}\left|\frac{5}{\sqrt{3}}\right|$