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

Question:

If a cuver passes through the point $(1,-2)$ and has slope of the tangent at any point $(\mathrm{x}, \mathrm{y})$ on it as

$\frac{x^{2}-2 y}{x}$, then the curve also passes through the point :

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

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

  3. $(-1,2)$

  4. $(3,0)$


Correct Option: , 2

Solution:

$\frac{d y}{d x}=\frac{x^{2}-2 y}{x}$ (Given)

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

$\mathrm{I} F=e^{\int \frac{2}{x} d x}=x^{2}$

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

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

hence bpasses through $(1,-2) \Rightarrow \mathrm{C}=-\frac{9}{4}$

Now check option(s), Which is satisly by option (ii)

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