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, the equation of the 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 = 4y

⇒ y + 4y = x

⇒ + 4 =

⇒ = – 4

We can see that it is a linear differential equation.

Comparing it with

P = , Q = – 4

I.F = e^∫Pdy

= e^dy

= e ^{– logy}

=

Solution of the given equation is given by

x × I.F = ∫Q × I.F dy + logc

⇒ x × () = ∫ – 4 × dy + logc

⇒ = – 4 log y + log c

⇒ = log y ^{– 4} + logc

⇒ = log c y ^{– 4}

⇒ = c y ^{– 4}