Find the equation of the tangent to the curve `(1+x^2)y=2-x ,` where it crosses the x-axis.

Options

A. `x+5y=2`
B. `x-5y=2`
C. `5x-y=2`
D. `5x+y=2`

Answer

A

Explanation

We have, equation of the curve `y(1+x^(2))=2-x`
`therefore y.(0+2x)+(1=x^(2)).(dy)/(dx)=0-1` [on differentiaing w.r.t. x]
`rArr 2xy+(1+x^(2))(dy)/(dx)=-1`
`rArr (dy)/(dx) = (-1-2xy)/(1+x^(2))`
Since, the given curve passes through X-axis i.e., y=0 [uisng Eq. (i)]
`therefore 0(1+x^(2))=2-x`
`rArr x=2`
So, the curve passes through the point (2,0).
`therefore (dy)/(dx)_(2,0)=(-1 -2 xx 0)/(1+2^(2))=-1/5`= Slope of the curve
`therefore` Slope of the tangent to the curve `=-1/5`
`therefore` Equation of tangent of the curve passing through (2,0) is
`y-0=-1/5(x-2)`
`rArr y+x/5=2/5`
`rArr 5y+x=2`