Question
The differential equation of family of curves whose tangent form an angle of $$\frac{\pi }{4}$$ with the hyperbola $$xy = {C^2}$$ is :
A.
$$\frac{{dy}}{{dx}} = \frac{{{x^2} + {C^2}}}{{{x^2} - {C^2}}}$$
B.
$$\frac{{dy}}{{dx}} = \frac{{{x^2} - {C^2}}}{{{x^2} + {C^2}}}$$
C.
$$\frac{{dy}}{{dx}} = - \frac{{{C^2}}}{{{x^2}}}$$
D.
none of these
Answer :
$$\frac{{dy}}{{dx}} = \frac{{{x^2} - {C^2}}}{{{x^2} + {C^2}}}$$
Solution :
Let the slope of tangent of required family be $$\frac{{dy}}{{dx}} = {m_1}$$
Also $$y = \frac{{{C^2}}}{x}\,;$$ therefore, $$\frac{{dy}}{{dx}} = - \frac{{{C^2}}}{{{x^2}}} = {m^2}\,\,\,\left( {{\text{say}}} \right)$$
By the given condition, we have
$$\eqalign{
& \tan \frac{\pi }{4} = \frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}} \cr
& \Rightarrow 1 + {m_1}{m_2} = {m_1} - {m_2} \cr
& \Rightarrow \frac{{dy}}{{dx}} + \frac{{{C^2}}}{{{x^2}}} = 1 - \frac{{{C^2}}}{{{x^2}}}\frac{{dy}}{{dx}} \cr
& \Rightarrow \frac{{dy}}{{dx}}\left( {1 + \frac{{{C^2}}}{{{x^2}}}} \right) = 1 - \frac{{{C^2}}}{{{x^2}}} \cr
& \Rightarrow \frac{{dy}}{{dx}} = \frac{{{x^2} - {C^2}}}{{{x^2} + {C^2}}} \cr} $$