Question

What is the slope of the tangent line of `sqrt(y-e^(x-y))= C`, where C is an arbitrary constant, at `(-2,1)`?

Answer 1

Equation of tangent is `y-1=1/(e^3+1)(x+2)`

Explanation:

As tangent is saught at the point `(-2,1)`, it is apparent that the point lies on the curve `sqrt(y-e^(x-y))=C`, we have

`sqrt(1-e^(-2-1))=C`

i.e. `C=sqrt(1-1/e^3)`

Hence function is `sqrt(y-e^(x-y))=sqrt(1-1/e^3)`

or `y-e^(x-y)=1-1/e^3`

As slope of tangent is the value of first derivative at the point, here `(-2,1)`, let us find first derivative by implicit differentiation. Differentiating `y-e^(x-y)=1-1/e^3`, we have

`(dy)/(dx)-e^(x-y)(1-(dy)/(dx))=0`

or `(dy)/(dx)(1+e^(x-y))=e^(x-y)`

i.e. `(dy)/(dx)=e^(x-y)/(1+e^(x-y))`

Hence, slope of tangent is `e^(-3)/(1+e^(-3))` or `1/(e^3+1)`

and equation of tangent is `y-1=1/(e^3+1)(x+2)`

graph{(sqrt(y-e^(x-y))-sqrt(1-1/e^3))(x-(e^3+1)y+3+e^3)=0 [-11.29, 8.71, -4.52, 5.48]}