Find the equation of the curve passing through the point (1, 1), if the perpendicular distance of the normal at P(x, y) to the curve from the origin is equal to the distance of P from the x - axis.

Here, P(x, y) is the point on the curve y = f(x)

We can find the equation of normal at P by

Here, The equation of normal at P is

- - - (1)

It is given that the distance from the origin is equal to the distance from the x - axis

Therefore,

…(i)

Let, F(x, y) =

Now, put x = λx and y = λy in F(x, y)

Taking λ as common from both numerator and denominator

We know, If the degree of the λ in function F(x, y) then it is said Homogenous function

So, the given differential equation is Homogenous differential equation.

Put y = vx in equation(i)

Differentiation y w.r.t x , we get

…(iii)

Now, Compare the equation (i) and (iii), we get

Taking x² as common from both numerator and denominator

Integrating both sides, we get

Let v² + 1 = t

On differentiating we get,

2v dv = dt, So

log t = - log x + log C

Putting the value of t , we get

Log|v² + 1| + log|x| = log|c|

log|x(v² + 1)| = log|C|

Now, putting back the value of v = y/x ,

- - - (iv)

And, The curve is passing through (1, 1)

Then, x = 1 and y = 1

(1)² + 1 = C(1)

C = 2

Put the value of C in equation (iv)

y² + 1 = x(2)

y² = 2x - 1

Hence, The curve is y² = 2x - 1