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Let slope of the tangent line to a curve at any point P(x, y) be given by $\frac{x y^{2} + y}{x} .$ If the curve intersects the line x + 2y = 4 at x = -2, then the value of y, for which the point (3, y) lies on the curve, is:

Option 1 -

$- \frac{4}{3}$

Option 2 -

$- \frac{18}{11}$

Option 3 -

$- \frac{18}{19}$

Option 4 -

$\frac{18}{35}$

2 Views | Posted a month ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

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

OR $\frac{d y}{d x} - \frac{y}{x} = y^{2} O R \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{x} . \frac{1}{y} = 1 . . . . . . . . . (i)$

=> $\frac{x}{y} = - \frac{x^{2}}{2} + c . . . . . . . . . . (i i)$

Since curve intersect x + 2y = 4 at x = -2 then y = 3 so

From (ii) $\frac{- 2}{3} = - 2 + c O R c = 2 - \frac{2}{3} = \frac{4}{3}$

put x = 3, then $\frac{3}{y} = \frac{- 9}{2} + \frac{4}{3} = \frac{- 19}{6}$

$\Rightarrow y = \frac{- 18}{19}$

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Let slope of the tangent line to a curve at any point P(x, y) be given by $\frac{x y^{2} + y}{x} .$ If the curve intersects the line x + 2y = 4 at x = -2, then the value of y, for which the point (3, y) lies on the curve, is:

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Let slope of the tangent line to a curve at any point P(x, y) be given by x y 2 + y x . If the curve intersects the line x + 2y = 4 at x = -2, then the value of y, for which the point (3, y) lies on the curve, is:

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Let slope of the tangent line to a curve at any point P(x, y) be given by $\frac{x y^{2} + y}{x} .$ If the curve intersects the line x + 2y = 4 at x = -2, then the value of y, for which the point (3, y) lies on the curve, is: