If a curve passes through the point (1,-2)


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

  1. (1) $(3,0)$

  2. (2) $(\sqrt{3}, 0)$

  3. (3) $(-1,2)$

  4. (4) $\quad(-\sqrt{2}, 1)$

Correct Option: , 2


$\because$ Slope of the tangent $=\frac{x^{2}-2 y}{x}$

$\because \quad \frac{d y}{d x}=\frac{x^{2}-2 y}{x}$

$\frac{d y}{d x}+\frac{2}{x} y=x$

I.F, $=e^{\int^{\frac{1}{4 x}}}=e^{2 \ln x}=x^{2}$

Solution of equation

$y \cdot x^{2}=\int x \cdot x^{2} d x$

$x^{2} y=\frac{x^{4}}{4}+C$

$\therefore$ curve passes through point $(1,-2)$


$\Rightarrow \quad C=\frac{-9}{4}$

Then, equation of curve

$y=\frac{x^{2}}{4}-\frac{9}{4 x^{2}}$

Since, above curve satisfies the point.

Hence, the curve passes through $(\sqrt{3}, 0)$.


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