Question

The curve that satisfies the differential equation $y^{\prime }=\frac{x^2+y^2}{2xy}$ and passes through (2, 1) is a hyperbola with eccentricity :

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $2$
• D. $\sqrt{5}$

Answer 1

Answer: (A)

Putting $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$, the given equation reduces to the form
$v+x\frac{dv}{dx}=\frac{x^2+v^2{x}^2}{2x\left(vx\right)}=\frac{1+v^2}{2v}\Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{2v}-v=\frac{1-v^2}{2v}$
$\Rightarrow \frac{2vdv}{1-v^2}=\frac{dx}{x}\Rightarrow -log\left(1-v^2\right)=logx-a$
$\Rightarrow log\left(1-v^2\right)x=a\Rightarrow x\left(1-v^2\right)=c$, where $c=e^a$
When $x=2,v=\frac{1}{2}$, [when $x=2y=1,v=\frac{1}{2}$]
so that $c=\frac{3}{2}$
So the equation of the curve is $x^2-y^2=\frac{3x}{2}$
$\Rightarrow {\left(x-\frac{3}{4}\right)}^2-y^2=\frac{9}{16}$
which is a rectangular hyperbola with eccentricity $\sqrt{2}$.
[Eccentricity of a rectangular hyperbola $=\sqrt{2}$ ]