Question

Question #2fcf6

Answer 1

ans.the equation is `y=x-1`.

Explanation:

The equation of the curve is `x+y-1=ln(x^2+y^2)`
we can write it as `e^(x+y-1)=x^2+y^2`
or,`e^(x+y)=e(x^2+y^2)`
or `e^x e^y=e(x^2+y^2)` `color (red) (1)`
now differentiating eq. 1 w.r.t `x` we get
`e^x e^y (dy/dx)` +`e^x e^y` =`2ex+2ey(dy/dx)`
rearranging we get ,`(dy/dx) =(2ex-e^(x+y))/(e^(x+y)-2ey)`

so we get `(dy/dx) at (1,0)` =`(2e-e)/(e-0)` =`e/e`=`1`.
so the eq. of the tangent line at the point`(1,0)` will be ,
`(y-0)=(dy/dx) at (1,0) (x-1)`
or `y=(x-1)`. (ans).

Answer 2

`y=x-1`

Explanation:

`x+y-1=ln(x^2+y^2)`. Term-by-term differentiation gives

`1+y'=1/(x^2+y^2)(2x+2yy')`. At P(1, 0) this becomes

1+y'=2, giving y'=0, at P.

So, the equation to the tangent at P)1, 0) is

`y-0=(1)(x-1)`

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]}