Question

Equation of the curve through (2, 1) whose slope at the point $(x, y)$ is $\frac{x^2+y^2}{2\,xy} is$

• A. $2 (x^2 - y^2) = 3x $
• B. $2 (y^2 - x^2) = 6y $
• C. $x (x^2 - y^2) = 6 $
• D. none of these.

Answer 1

Answer: (A)

$\frac{dy}{dx} = \frac{x^{2}+y^{2}}{2xy}$
$Put y = vx$
$\therefore v+x \frac{dv}{dx} = \frac{1+v^{2}}{2v}$
$\Rightarrow x \frac{dv}{dx} = \frac{1+v^{2}}{2v}$
$\Rightarrow x \frac{dv}{dx} = \frac{1+v^{2}}{2v}-v = \frac{1-v^{2}}{2v}$
$\Rightarrow \frac{2v}{1-v^{2}} dv = \frac{dx}{x}
\Rightarrow \int -\frac{2v}{1-v^{2}} dv +\int \frac{dx}{x} = const.$
$\Rightarrow log \left(1 - v^{2}\right) + log\,x = log \,C$
$\Rightarrow log\,x\left(1- \frac{y^{2}}{x^{2}} \right) = log\,C$
$\Rightarrow x^{2} -y^{2} = Cx$
$It\, passes\, through \left(2, 1\right)$
$\therefore 4-1 = 2C$
$\therefore C = \frac{3}{2}$
$\therefore x^{2}-y^{2} = \frac{3}{2} x$
$\Rightarrow 2 \left(x^{2}-y^{2}\right) = 3x$