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$The x-intercept of the tangent to a curve is equal to the coordinate of the point of contact.$
$The equation of the curve through the point (1,1) is$

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a

yex/y=e

b

xex/y=e

c

xey/x=e

d

yey/x=e

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detailed solution

Correct option is A

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−log⁡V=log⁡x+c⇒−x/y−log⁡y+log⁡x=log⁡x+c⇒−xy=log⁡y+c pass (1,1)⇒c=−1⇒y=e⋅e−x/y⇒ex/y=ey⇒yex/y=e

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