Question

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

• 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)

$\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$
$Puty=vx$
$\therefore v+x\frac{dv}{dx}=\frac{1+v^2}{2v}$
$\Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{2v}$
$\Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{2v}-v=\frac{1-v^2}{2v}$
$\Rightarrow \frac{2v}{1-v^2}dv=\frac{dx}{x}<br>\Rightarrow \int -\frac{2v}{1-v^2}dv+\int \frac{dx}{x}=const.$
$\Rightarrow log\left(1-v^2\right)+logx=logC$
$\Rightarrow logx\left(1-\frac{y^2}{x^2}\right)=logC$
$\Rightarrow x^2-y^2=Cx$
$Itpassesthrough\left(2,1\right)$
$\therefore 4-1=2C$
$\therefore C=\frac{3}{2}$
$\therefore x^2-y^2=\frac{3}{2}x$
$\Rightarrow 2\left(x^2-y^2\right)=3x$