Question

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

Answer 1

Answer: (B)

$\frac{dy}{dx}=\frac{x^2-2Y}{x}(Given)$
$\frac{dy}{dx}+2\frac{Y}{x}=x$
$I.F=e^{\int \frac{2}{x}dx}=x^2$
$\therefore y.x^2=\int x.x^{2}dx+C$
$=\frac{x^4}{y}+C$
hence bpasses through (1, -2) $\Rightarrow C=-\frac{9}{4}$
$\therefore y{x}^2=\frac{x^4}{4}-\frac{9}{4}$
Now check option(s) , Which is satisly by option (ii)