Find the equation of the curve such that the portion of the x - axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).
Let P(x,y) be the point of contact of tangent and curve y = f(x).
It cuts the axes at A and B so, equation of tangent at P(x,y)
Y – y = (X – x)
Putting X = 0
Y – y = (0 – x)
⇒ Y = y – x
So, A(0, y – x)
Now, putting Y = 0
0 – y = (X – x)
⇒ X = x – y
So, B(x – y,0)
Given, intercept on x - axis = 4× ordinate
⇒ x – y = 2x
⇒ – y = x
⇒
⇒ – logx = logy + c ……(1)
As it passes through (1,2)
So, the point must satisfy the equation above
– log1 = log2 + c
⇒ 0 = log2 + c
⇒ c = – log2
Putting the value of c in equation (1)
– logx = logy – log2
⇒ log2 = logx + logy
⇒ log2 = logxy
⇒ xy = 2