1/f=1/v+1/uHere u is the object distance v is the image diatnceLet us take the case of concave mirror.   See diagram.Let P be the pole (geometrical center) of the mirror.  Let us draw the axis of the mirror.  Let us draw the focal point F and the center C at distances f and 2f from the pole on the principal axis.Let us take a point object at A which is at distance u from P.  To find its image draw a ray AM on to the mirror at angle y to the axis.  Draw the normal to the mirror at M which passes through the center C.  Now draw the reflected ray making the same angle x with the normal.  Let it intersect the axis at A'.  Draw another ray from object A along the axis which gets reflected back along the same path.  It intersects A'M at A'.   Hence A' is the image of A.We want to find a relation between u and v (image distance from P),Z = exterior angle of Triangle AMA' and hence  z = y + 2 x  or,  z - 2x = yz-x is the exterior angle of triangle AMC.Tan y = MP' / AP'           Tan z =  MP' / A'P'         tan z-x = MP' / CP'When y, x, z are all small angles.  Tangent is nearly same as the angle.  Hence         y = MP' / AP'            z =  MP' / A'P'          z-x = MP' / CP'So we  get  x = MP'/AP' - MP'/CP'Since y = z - 2x ,    MP' / AP' = MP' / A' P' - 2 [  MP'/AP' - MP'/CP' ]    Cancel MP'.             1 /AP' = 1 / A' P' - 2 / A'P' + 2 /CP'Substitute CP' = 2f and A'P' = v   and u = AP'.   as  PP' is very small and is negligible for small aperture mirror and small angles x, y and z.           1 / u = - 1 / v + 2 / 2f            1/u + 1/v = 1/f   -- equation 1We introduce a cartesian sign convention. The distances in the direction of the light ray are taken as positive and those in the opposite direction are taken as negative.So u is -ve,  v is negative and f is also negative.  Then  equation 1 is          1/-u  + 1/-v  = 1/-f  =>     1/u + 1/v = 1/fThe same equation can be proved for convex mirror also.

But I want it's proof