Let slope of the tangent line to a curve at any point P(x, y) be given by [xy^2 + y x].

Question

Let slope of the tangent line to a curve at any point P(x, y) be given by $${{x{y^2} + y} \over x}$$. 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 :

Accepted Answer

$${{dy} \over {dx}} = {{x{y^2} + y} \over x}$$

$$ \Rightarrow {{xdy - ydx} \over {{y^2}}} = xdx$$
$$ \Rightarrow  - d\left( {{x \over y}} \right) = d\left( {{{{x^2}} \over 2}} \right)$$

$$ \Rightarrow {{ - x} \over y} = {{{x^2}} \over 2} + C$$

Curve intersect the line x + 2y = 4 at x = - 2

So, - 2 + 2y = 4 $$ \Rightarrow $$ y = 3

So the curve passes through (-2, 3)

$$ \Rightarrow {2 \over 3} = 2 + C$$

$$ \Rightarrow C = {{ - 4} \over 3}$$

$$ \therefore $$ curve is $${{ - x} \over y} = {{{x^2}} \over 2} - {4 \over 3}$$

It also passes through (3, y)

$${{ - 3} \over y} = {9 \over 2} - {4 \over 3}$$

$$ \Rightarrow {{ - 3} \over y} = {{19} \over 6}$$

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

Solutions · 1