Question #2fcf6
2 Answers
ans.the equation is
Explanation:
The equation of the curve is
we can write it as
or,
or
now differentiating eq. 1 w.r.t
rearranging we get ,
so we get
so the eq. of the tangent line at the point
or
Explanation:
1+y'=2, giving y'=0, at P.
So, the equation to the tangent at P)1, 0) is
Note in the graphs the tangent crossing the curve at the point of
inflexion P(1, 0). The scales, for the two graphs, are different
graph{(x+y-1-ln(x^2+y^2))(y-x+1)((x-1)^2+y^2-.01)=0 [-5, 5, -2.5, 2.5]}
graph{(x+y-1-ln(x^2+y^2))(y-x+1)((x-1)^2+y^2-.0001)=0 [0, 2, -.5, .5]}