JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 3)
If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as $${{{x^2} - 2y} \over x}$$, then the curve also passes through the point :
(–1, 2)
$$\left( { - \sqrt 2 ,1} \right)$$
$$\left( { \sqrt 3 ,0} \right)$$
(3, 0)
Explanation
$${{dy} \over {dx}} = {{{x^2} - 2y} \over x}$$ (Given)
$${{dy} \over {dx}} + 2{y \over x} = x$$
I.F = $${e^{\int {{2 \over x}dx} }} = {x^2}$$
$$ \therefore $$ y.x2 = $$\int {x.{x^2}} dx + C$$
= $${{{x^4}} \over y} + C$$
This curve passes through (1, $$-$$ 2) $$ \Rightarrow $$ C= $$-$$ $${9 \over 4}$$
$$ \therefore $$ yx2 = $${{{x^4}} \over 4} - {9 \over 4}$$
Now check option(s), Which is satisly by option (ii)
$${{dy} \over {dx}} + 2{y \over x} = x$$
I.F = $${e^{\int {{2 \over x}dx} }} = {x^2}$$
$$ \therefore $$ y.x2 = $$\int {x.{x^2}} dx + C$$
= $${{{x^4}} \over y} + C$$
This curve passes through (1, $$-$$ 2) $$ \Rightarrow $$ C= $$-$$ $${9 \over 4}$$
$$ \therefore $$ yx2 = $${{{x^4}} \over 4} - {9 \over 4}$$
Now check option(s), Which is satisly by option (ii)
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