Answer to Question #16315 in Calculus for hsd
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).
1
2012-10-11T11:25:00-0400
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.
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