Question

If a curve passes through the point $(1,-2)$ and has slope of the tangent at any point $(x,y)$ on it as $\frac{x^2-2y}{x}$, then the curve also passes through the point :

• A. $(3,0)$
• B. $(-1,2)$
• C. $(-\sqrt{2},1)$
• D. $(\sqrt{3},0)$

Answer 1

The correct option is D $(\sqrt{3},0)$

Slope of tangent $=\frac{x^2-2y}{x}$
$\therefore \frac{dy}{dx}=x-\frac{2y}{x}$
$\Rightarrow \frac{dy}{dx}+\frac{2}{x}y=x$
$I.F.=\exp \left(\int \frac{2}{x}dx\right)=\exp \left(2\log x\right)=x^2$

$\therefore$ The general solution is
$x^{2}y=\int x^{3}dx$
$\Rightarrow x^{2}y=\frac{x^4}{4}+C$
$\because$ The curve passes through $(1,-2)$
$\therefore -2-\frac{1}{4}=C\Rightarrow C=-\frac{9}{4}$
$\therefore$ The curve is $y{x}^2=\frac{x^4}{4}-\frac{9}{4}$
$(\sqrt{3},0)$ lies on the above curve.