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Find the equation of the tangent to the curve `(1+x^2)y=2-x ,`where it crosses the x-axis.A. `x+5y=2`B. `x-5y=2`C. `5x-y=2`D. `5x+y=2` |
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Answer» Correct Answer - A 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` |
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