Question
The differential equation $$\left( {1 + {y^2}} \right)x\,dx - \left( {1 + {x^2}} \right)y\,dy = 0$$ represents a family of :
A.
ellipses of constant eccentricity
B.
ellipses of variable eccentricity
C.
hyperbolas of constant eccentricity
D.
hyperbolas of variable eccentricity
Answer :
hyperbolas of variable eccentricity
Solution :
Given $$\frac{{x\,dx}}{{1 + {x^2}}} = \frac{{y\,dy}}{{1 + {y^2}}}$$
Integrating we get,
$$\eqalign{
& \frac{1}{2}\log \left( {1 + {x^2}} \right) = \frac{1}{2}\log \left( {1 + {y^2}} \right) + a \cr
& \Rightarrow 1 + {x^2} = c\left( {1 + {y^2}} \right),\,c = {e^{2a}} \cr
& {x^2} - c{y^2} = c - 1 \Rightarrow \frac{{{x^2}}}{{c - 1}} - \frac{{{y^2}}}{{\left( {\frac{{c - 1}}{c}} \right)}} = 1......\left( 1 \right) \cr} $$
Clearly $$c > 0{\text{ as }}c = {e^{2a}}$$
Hence, the equation $$\left( 1 \right)$$ gives a family of
hyperbolas with eccentricity
$$ = \sqrt {\frac{{c - 1 + \frac{{c - 1}}{c}}}{{c - 1}}} = \sqrt {\frac{{{c^2} - 1}}{{c - 1}}} = \sqrt {c + 1} {\text{ if }}c \ne 1$$
Thus eccentricity varies from member to member of the family as it depends on $$c.$$ If $$c = 1,$$ it is a pair of lines $${x^2} - {y^2} = 0$$