If a curve passes through the origin and the slope of the tangent to it at any point (x,y) is x2−4x+y+8x−2, then this curve also passes through the point :
A
(4,5)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(5,4)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(4,4)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(5,5)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D(5,5) dydx=(x−2)2+y+4(x−2)=(x−2)+y+4(x−2)
Let x−2=t⇒dx=dt
and y+4=u⇒dy=dudydx=dudtdudt=t+ut⇒dudt−ut=tI.F.=e∫−1tdt=e−lnt=1tu×1t=∫t×1tdt⇒ut=t+cy+4x−2=(x−2)+c
Passing through (0,0) c=0 ⇒(y+4)=(x−2)2
It passes though (5,5)