9 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Is there a figure too?
My sir said you can make figure by own
Although this is the given one
We didn't know P or T. 😅
(Neither are they standard notation)
The equation for the tangent is y-y_0=y'(x-x0), from this, find the x and y intercept and use midpoint formula to obtain the required d.e.
After that solve that to prove the curve is indeed a parabola
I'm writing this a bit directly but:
$\frac{\frac{-y{0}+x{0}y'}{y'}+x{0}}{2}=0 \newline \frac{y{0}}{2}=-y'.x{0}+y_{0}$
SirLancelotDuLac
Replace x_0 and y_0 by x and y in hindsight
(For clarity in notation)
From the first equation itself we get 2xy'=y
Which means 2dy/y=dx/x
Which means 2lny=ln(x)+c
Which means x=cy^2
@Abhi You there?
Ohk
Phone got discharged