a tangent and the normal to a curve at the point (x,y) makes equal intercepts on the x axis and the y axis respectively find the curve which passes through point (1,1)
So, what do we know?
1) We've a curve. y = f(x)
2) We've two lines. y = mx + b (Tangent) y = nx + c (Normal)
3) They all three intersect at the given arbitrary point (s,t): f(s) = m(s) + b = n(s) + c = t
4) Two slopes are equal at (x,y): f'(s) = m
5) Definition of Normal: m*n = -1
6) Intercepts: 0 = m(x) + b (Tangent) y = n(0) + c (Normal) ==> c = -b/m
There's a lot of information in there. Let's see if you get all the same information and what you can do with it.
Equivalently: the equations of the tangent and normal lines to $\displaystyle y=y(x)$ at $\displaystyle (x,y)$ are respectively $\displaystyle Y-y=y'(X-x)$ , $\displaystyle Y-y=(-1/y')(X-x)$ . For $\displaystyle Y=0$ and $\displaystyle X=0$ and making the corresponding segments equal you'll obtain a differential equation. Find the particular solution satisfying $\displaystyle y(1)=1$ .