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The x-intercept of the tangent to a curv...

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

A

`ye^(x//y)=e`

B

`xe^(x//y)=e`

C

`xe^(y//x)=e`

D

`ye^(y//x)=e`

Text Solution

Verified by Experts

The correct Answer is:
A

Equation of tangent is `Y-y=(dy)/(dx)(X-x)`
For x-intercept, Y=0 . Thus, `X=x-y(dx)/(dy)`.
According to question, `x-y(dx)/(dy)=y`
or `(dy)/(dx)=y/(x-y)`
Putting `y=vx`, we get
`v+x(dv)/(dx)=v/(1-v)`
or `x(dv)/(dx)=v/(1-v)-v=(v-v+v^(2))/(1-v)`
or `int(1-v)/(v^(2))dv=int(dx)/x`
or `-x/y-logy/x=logx+c`
Given when `x=1, y=1`, then `c=-1`
Hence, equation of curve is `1-x/y=logy`
or `y=ee^(-x//y)`
or `ye^(x//y)=e`
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