# Answer to Question #16315 in Calculus for hsd

Question #16315

Determine the equations of the tangent lines and of the normal lines to the curve

e^(x^2+y^2-2) = x

at the points (1, 1) and (1,-1).

e^(x^2+y^2-2) = x

at the points (1, 1) and (1,-1).

Expert's answer

1. We find derivative of this curve.

d/dx(e^(x^2+y^2)-x)=0

(e^(x^2))*(e^(y^2))'+(e^(x^2))' * (e^(y^2))=1

So, y'y+x=0.5e^-(x^2+y^2)

y'=(0,5e^x^2-y^2-x)/y

Tangent line equation:

(x-x0)/(y-y0)=y'(x0)

Normal:

(x-x0)/(y-y0)=-y'(x0)

In point (1;1)

(x-1)=(0.5(e^-2)-1)(y-1)

Normal: (x-1)=(0.5(e^-2)-1)(1-y)

in point (1;-1)

Tangent:

(x-1)=-(y+1) =>x+y=0

Normal:

x-y=0.

d/dx(e^(x^2+y^2)-x)=0

(e^(x^2))*(e^(y^2))'+(e^(x^2))' * (e^(y^2))=1

So, y'y+x=0.5e^-(x^2+y^2)

y'=(0,5e^x^2-y^2-x)/y

Tangent line equation:

(x-x0)/(y-y0)=y'(x0)

Normal:

(x-x0)/(y-y0)=-y'(x0)

In point (1;1)

(x-1)=(0.5(e^-2)-1)(y-1)

Normal: (x-1)=(0.5(e^-2)-1)(1-y)

in point (1;-1)

Tangent:

(x-1)=-(y+1) =>x+y=0

Normal:

x-y=0.

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