Question-40 The angle between the curves xy=6 and x^2y=12 is
(A) \tan^{-1}\frac{3}{11}
(B) \tan^{-1}\frac{11}{3}
(C) \tan^{-1}\frac{2}{11}
(D) \tan^{-1}\frac{1}{11}
Answer: (A)
Explanation:
Solving xy=6 and x^2y=12, we get
x=2,\ y=3∴ Consider, xy=6
⇒ y=\frac{6}{x}
⇒ \frac{dy}{dx}=-\frac{6}{x^2}
m_1=\left(\frac{dy}{dx}\right)_{x=2}=\frac{-6}{4}=-\frac{3}{2}Now, x^2y=12
⇒ y=\frac{12}{x^2}
⇒ \frac{dy}{dx}=\frac{12(-2)}{x^3}=-\frac{24}{x^3}
m_2=\left(\frac{dy}{dx}\right)_{x=2}=\frac{-24}{8}=-3Using, \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|
⇒ \theta=\tan^{-1}\left|\frac{m_1-m_2}{1+m_1m_2}\right|
=\tan^{-1}\left|\frac{-\frac{3}{2}+3}{1+\left(-\frac{3}{2}\right)(-3)}\right| =\tan^{-1}\left(\frac{3}{11}\right)