The x-intercept of the tangent to a curve is equal to the coordinate of the point of contact.
The equation of the curve through the point (1,1) is
yex/y=e
xex/y=e
xey/x=e
yey/x=e
equation of tangent Y−y=dydx(x−x) For X-intercept Y=0→x=x−ydxdy According to the question x−ydxdy=y→dydx=yx−y put y=Vx→xdvdx=v21−v→∫1−vv2dv=∫dxx⇒−1V−logV=logx+c
⇒−x/y−logy+logx=logx+c⇒−xy=logy+c pass (1,1)⇒c=−1⇒y=e⋅e−x/y⇒ex/y=ey⇒yex/y=e