JEE Main 25 Feb 2021 Shift 1 Solved Paper

Given, slope =‌

x2−4x+y+8

x−2

⇒‌‌‌

dy

dx

‌‌=‌

x2−4x+y+8

x−2

=‌

(x−2)2+(y+4)

(x−2)

‌‌=(x−2)+‌

y+4

x−2

Let (x−2)=t⇒dx=dt
‌ and ‌(y+4)=u⇒dy=du
∴‌‌‌

dy

dx

=‌

du

dt

‌ Now, ‌‌

dy

dx

=(x−2)+‌

(y+4)

(x−2)

⇒‌‌‌

du

dt

=t+‌

u

t

⇒‌

du

dt

−‌

u

t

=t
Here, integrating factor (IF) =1∕t
⇒‌‌u⋅(‌

1

t

)=∫t(‌

1

t

)dt⇒u∕t=t+c
⇒‌‌‌

(y+4)

(x−2)

=(x−2)+c
∵ It passes through origin [i.e. (0,0) ], then
∴‌‌‌

4

−2

=−2+c
⇒‌‌−2+2=c⇒c=0
‌ Hence, ‌‌

(y+4)

(x−2)

=(x−2)+0
⇒‌‌y+4=(x−2)2
Clearly, this curve passes through (5,5) as it satisfies the equation.