Question

The angle at which the curve $y=x^2$ and the curve $x=\frac{5}{3}\cos t,y=\frac{5}{4}\sin t$ intersect is :

• A. $2{\tan }^{-1}\frac{41}{2}$
• B. ${\tan }^{-1}\frac{2}{41}$
• C. ${\tan }^{-1}\frac{41}{2}$
• D. $-{\tan }^{-1}\frac{2}{41}$

Answer 1

The correct option is C ${\tan }^{-1}\frac{41}{2}$

$x=\frac{5}{3}\cos t,y=\frac{5}{4}\sin t$
$\Rightarrow 9x^2+16y^2=25$
$y=x^2$
The given two curves intersect at $(\pm 1,1)$

For $9x^2+16y^2=25$
$m_1=\frac{dy}{dx}{|}_{(1,1)}=\frac{-9}{16}$

For $y=x^2$
$m_2=\frac{dy}{dx}{|}_{(1,1)}=2$
$\Rightarrow \tan \alpha =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\frac{41}{2}\Rightarrow \alpha ={\tan }^{-1}\frac{41}{2}$