JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, slope of tangent line to curve at (x,y) is ‌

xy2+y

x

i.e., ‌‌‌

dy

dx

=‌

xy2+y

x

⇒‌‌

dy

dx

=y2+‌

y

x

⇒xdy=xy2dx+ydx
⇒‌xdy−ydx=xy2dx⇒‌‌

xdy−ydx

y2

=xdx
Integrating both sides, we get
‌

−x

y

=‌

x2

2

+C . . . (i)
The curve intersect line at x=−2
Then, x=−2, is satisfied by x+2y=4
Hence, (−2)+2y=4 Gives, y=3
∴ Curve passes through (2,−3).
Use (2,−3) is Eq. (i), we get
‌

−2

−3

=‌

(−2)2

2

+C⇒C=‌

−4

3

∴ The curve is
‌

−x

y

=‌

x2

2

=‌

−4

3

. . . (ii)
It also passes through (3,y).
Put (3,4) in Eq. (ii), we get
⇒‌‌‌

−3

y

=‌

(3)2

2

−‌

4

3

⇒y=−‌

18

19