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)

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