Y=(x^2) +1 and y = - (x^2)? tangent

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The discussion focuses on finding the equations of tangent lines that touch both curves y = (x^2) + 1 and y = - (x^2) simultaneously. It is established that the slopes of the tangents at the points of tangency must be equal, leading to the relationship m = 2x0 = -2x1, which implies x1 = -x0. By setting the equations for the y-intercepts equal, b = 1 - x0^2 and b = x1^2, it is determined that 1 - x0^2 = x0^2. Solving this equation reveals two symmetric solutions for x0, which can then be used to find the slopes and y-intercepts of the tangent lines. The complexity of the problem is acknowledged, highlighting the need for careful calculations.
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Find the equations of the lines that are tangent to both curves simultaneously:y=(x^2) +1 and y = - (x^2)? :eek:
 
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Find the equations for the tangent lines to both curves and set them equal to each other. You will find when the slope of the tangents are equal and then can make an equation(s) of of it.
 
Jameson is correct but it's a bit more complicated than he implies.

Suppose a line is tangent to y= x2+ 1 at (x0,x02+ 1) and tangent to y= -x2 at (x1,-x12).
Any (non-vertical) line can be written as y= mx+ b. m is equal to the derivative of the functions at the given points: m= 2x0= -2x1 so x1= -x0. We must have x02+1= (2x0)x0+ 1 or b= 1- x02. We must also have -x12= (-2x1)x1+ b or b= x12. That is, b= x12= 1- x02. But since x1= -x0, x12= x02 so 1- x02= x02.

Solve that for x0 and then you can find m and b.

Because of the squares, there are, of course, two symmetric solutions,.
 
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