JEE Main 12 Jan 2019 Evening Question 20
Question: 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}-2y}{x}, $ then the curve also passes through the point
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ (-\sqrt{2},1) $
B) $ (-1,2) $
C) $ (\sqrt{3},0) $
D) $ (3,0) $
Show Answer
Answer:
Correct Answer: C
Solution:
- Here, $ =\frac{dy}{dx}=\frac{x^{2}-2y}{x} $
$ \Rightarrow $ $ \frac{dy}{dx}+\frac{2}{x}y=x $
$ \therefore $ $ I.F.={e^{\int _{{}}^{{}}{\frac{2}{x}dx}}}=x^{2} $
$ \therefore $ Solutions is $ y.x^{2}=\int _{{}}^{{}}{x.x^{2}dx+C} $
$ \Rightarrow $ $ yx^{2}=\frac{x^{4}}{4}+C $
Since, the curve passes through the point ( $ 1,-2 $ ).
$ \therefore $ $ -2=\frac{1}{4}+C\Rightarrow C=\frac{-9}{4} $
$ \therefore $ $ yx^{2}=\frac{x^{4}}{4}-\frac{9}{4} $
$ \Rightarrow $ $ x^{4}-4x^{2}y-9=0 $
Now, only the point in option i.e., $ (\sqrt{3},0) $ satisfies the above equation.
