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

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The discussion focuses on finding the equations of the lines that are tangent to both curves defined by the equations y = (x^2) + 1 and y = - (x^2). The key approach involves setting the slopes of the tangents equal, leading to the relationship m = 2x0 = -2x1, which simplifies to x1 = -x0. By substituting this relationship into the equations for the tangent lines, the values of b can be expressed as b = 1 - x0^2 and b = x1^2. Solving these equations reveals two symmetric solutions for x0.

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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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