The angle between the curves xy=6 and x^2y=12 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given curves are $xy=6$ $(1)$ and $x^2y=12$ $(2)$. Dividing $(2)$ by $(1)$,we get $\frac{x^2y}{xy} = \frac{12}{6}$,which implies $x=2$. Substituti

Vedclass · Accepted Answer

$	an^{-1} \frac{3}{11}$

Vedclass · Answer

(A) Given curves are $xy=6$ $(1)$ and $x^2y=12$ $(2)$. Dividing $(2)$ by $(1)$,we get $\frac{x^2y}{xy} = \frac{12}{6}$,which implies $x=2$. Substituting $x=2$ in $(1)$,we get $2y=6$,so $y=3$. The point of intersection is $(2, 3)$. For curve $(1)$,$y = \frac{6}{x}$,so $\frac{dy}{dx} = -\frac{6}{x^2}$. At $(2, 3)$,$m_1 = -\frac{6}{4} = -\frac{3}{2}$. For curve $(2)$,$y = \frac{12}{x^2}$,so $\frac{dy}{dx} = -\frac{24}{x^3}$. At $(2, 3)$,$m_2 = -\frac{24}{8} = -3$. The angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. $	an 	heta = |\frac{-1.5 - (-3)}{1 + (-1.5)(-3)}| = |\frac{1.5}{1 + 4.5}| = |\frac{1.5}{5.5}| = \frac{15}{55} = \frac{3}{11}$. Therefore,$	heta = 	an^{-1} \frac{3}{11}$.

The angle between the curves $xy=6$ and $x^2y=12$ is

Similar Questions

If the length of the subnormal at any point on the curve $y = a^{1-n}x^n$ is constant,then $n = \dots$

If the slope of the tangent to the curve $y=ax^3+bx+4$ at the point $(2, 14)$ is $21$, then the values of $a$ and $b$ are respectively:

If the normal to the curve ${y^2} = 5x - 1$ at the point $(1, -2)$ is of the form $ax - 5y + b = 0$,then $a$ and $b$ are:

The tangent to the curve $y = xe^{x^2}$ passing through the point $(1, e)$ also passes through the point

Consider $f(x) = \tan^{-1}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right)$,where $x \in (0, \frac{\pi}{2})$. $A$ normal to $y = f(x)$ at $x = \frac{\pi}{6}$ also passes through the point:

Vedclass Test Series

Exam Paper Generator

Online Exam Module