IIT JEE Advanced 2006 Paper 1

The equation of tangent to any curve at point P(x, y) is
Y - y =

dy

dx

(X−x)
which cuts the x-axis at Y = 0:
X = x - y

dy

dx

that is, at
A (x−y

dy

dx

,0)
and it also cuts the y-axis at X = 0:
Y = y - x

dy

dx

That is, at
B (0,y−x

dy

dx

)

x =

[(x−y dy dx )×1]+(3×0)

1+3

4x = x - y

dy

dx

⇒ y

dx

dy

= - 3x
y =

(1×0)+[3(y−x dy dx )]

4

⇒ 4y = 3y - 3x

dy

dx

⇒ 3x

dy

dx

= - y
The equation of the curve is
3x

dy

dx

+ y = 0
3‌∫

dy

y

+∫

dy

x

= ln c
3lny + lnx = lnc
xy3 = c
which is passing through the point (1, 1). Therefore, c = 1 and hence
xy3 = 1
which is the equation of curve that passes through (1/8, 2).