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# at what point is the tangent to the curve x+y^2=1 parallel to the line x+2y=0

10 years ago

when the tangent to the curve x+y2=1 is parallel to another curve ,

then that means the derivative of this equation is equal to the derivative of the other curve i.e. x+2y =0

therefore, the derivative of the first curve is 1+ (2y)dy/dx=0

simplifying it we get dy/dx=-1/2y--------------(i)

now derivative of the other curve is 1+ 2dy/dx=0

simplifying it we get dy/dx=-1/2---------------(ii)

now if we compare these two equation we would see that we get y=1 [from -1/2y=-1/2]

at the point when y is equal to 1 x is equal to 0 . This we can get from the equation of the curve ......

so, the point where the curve x+y2=1 is parallel to the curve x+2y=0 is (0,1)

hope this helps....;)

10 years ago

x+y^2=1, upon implicit differentiation we get 1=-2yy',y'=-1/2y=slope of line=-(1/2),thus -1/2=-1/2y,y=+1,x=1-y^2=1-1=0

therefore the point is (0,1)