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

Answer 1

Answer: (B)

${\tan }^{-1}\frac{41}{2}$

Given
$y=x^2...\left(i\right)$
$x=\frac{5}{3}\cos t;y=\frac{5}{4}\sin t...\left(ii\right)$
Which is parametric equation, we change this equation is caresian equation as follows
$\cos t=\frac{3}{5}x;\sin t=\frac{4}{5}y$
On Squaring and adding both i.e., $\cos t$ and $\sin t$ we get
$\frac{9}{25}{x}^2+\frac{16}{25}{y}^2={\cos }^{2}t+{\sin }^{2}t$
$\Rightarrow 9x^2+16y^2=\mathrm{25....}\left(iii\right)[\because {\cos }^2\theta +{\sin }^2\theta =1]$
$\therefore$ The intersection points at Eq.(i) and (iii) are $(1,1)$ and $(-1,1)$
Now, slope of tangent of Eq.(i) at point $(1,1)$ is
$m_1=\frac{dy}{dx}=2x\therefore m_1={\left|\frac{dy}{dx}\right|}_{(1,1)}=2$
And slope of tangent of Eq. (iii) at point $(1,1)$ is
$m_2=\frac{dy}{dx}=-\frac{9}{16}$
$\therefore$ Angle at point of intersection of Eqs. (i) and (iii) we get
${\theta }_1={\tan }^{-1}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|={\tan }^{-1}\frac{41}{2}$
similarly, slope of tangent of Eq. (i) at point $(-1,1)$
$m_1={\left|\frac{dy}{dx}\right|}_{(-1,1)}=-2$
And slope of tangent of Eq. (iii) at point $(-1,1)$
$m_2=\frac{dy}{dx}=\frac{9}{16}$
$\therefore$ Angle at point of intersection of Eqs. (i) and (iii) we get
${\theta }_2={\tan }^{-1}\left|\frac{-2-\frac{9}{16}}{1-\frac{18}{16}}\right|={\tan }^{-1}\frac{41}{2}$