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).