Question

Consider the curve which satisfies the differential equation $(1+y^2)dx-xydy=0$ and also passes through the point (1,0) If the point (a, b), a > 0, is the focus of the curve, then the value of is equal to

Answer 1

From given differential equation
$\int \frac{dx}{x}=\int \frac{ydy}{1+y^2}$
$\Rightarrow \ln x=\frac{1}{2}\ln (1+y^2)+c$
Which passes through (1, 0)
$\therefore 0=\frac{1}{2}\ln (1+0)+c\Rightarrow c=0$
$\therefore \ln x=\frac{1}{2}\ln (1+y^2)$
$\Rightarrow x^2=1+y^2\Rightarrow x^2-y^2=1$ which is a rectangular hyperbola with foci $(\pm \sqrt{2},0)=(a,b)\Rightarrow 16(\sqrt{2}a+b)=32$