Let `y=f(x)` be a curve passing through `(1,1)` such that the triangle formed by the coordinate axes and the tangent at any point of the curve lies in the first quadrant and has area 2. Form the differential equation and determine all such possible curves.

Answer

Differential Equation: `(d^(2)y)/(dx^(2))=0, x^(2) dy/dx + 1 = 0`
curvers: `x+ y = 2, xy =1`

Explanation

Equation of tangent to the curve y= f(x) at point
`A(x,y) is Y-y=dy/dx(X-x)`
whose, x-intercept `(x-y cdot dx/dy, 0)`
y-intercept `(0,y-xdy/dx)`
Given, `Delta OPQ=2`
`rArr 1/2 (x-ydx/dy)(y-xdy/dx)=2`
`rArr (x-y 1/p)(y-xp)=4, where p=dy/dx`
`rArr p^(2) x^(2)-2pxy +4p+y^(2)=0`
`rArr (y-px)^(2) + 4p=0`
`therefore y-px =2 sqrt(-p)`
`rArr y=px+2 sqrt(-p) …(i)`
on differentiating w.r.t.x, we get
`p=p+(dp)/dx cdot x+ 2 cdot (1/2)(-P) ^(1//2) cdot (-1) (dp)/dx`
`rArr (dp)/dx{x-(-P)^(1//2)}=0`
`rArr (dp)/dx= 0 or x=(-p)^(-1) `
If `(dp)/dx= 0 rArr P=c`
On putting this value in Eq. (i), we get `y=cx+2sqrt(-c)`
This curve passes through (1,1).
`rArr 1=c+2sqrt(-c)`
`rArr c=-1`
`therefore y=-x+2`
`rArr x+y= 2 `
Again, if `x=(-p)^(-1//2)`
`rArr -p=1/x^(2)` putting in Eq. (i)
`y=(-x)/x^(2) + 2 cdot 1/x rArr xy=1`
Thus, the two curves are xy = 1 and x+ y = 2.