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