Find the two common tangents of y=-x^2 and y=Ax^2+C?

Consider the parabolas y=x^2 and y=Ax^2+C, where A and C are positive constants. FOr a given choice of A and C there will be exactly two common tangents (each line is tangent to both parabolas), symmetrically placed with respect to the y-axis.





a) Find the points of tangency on each parabola and the equations of the common tangents, each expressed in terms of A and C.





b) Find the points and tangent lines for the particular values A=3 and B=12, for the parabola y=3x^2+12.











To anyone who could solve this..thank you and God bless you. %26gt;_%26lt;

Find the two common tangents of y=-x^2 and y=Ax^2+C?
The two eqns f(x) = x^2 and g(x) = Ax^2 + C are clearly parabolas.





f(x) has its vertex at (0, 0) and g has vertex (0, C). Therefore f and g intersect only if g is “flat” otherwise f would always be on the outside of g. That is, the coefficient of x^2 in fcn f(x) is 1; as a result, the only way these two fcns could ever intersect is if A %26lt; 1. I.e. g(x) is “flat”.





Furthermore, the graphs share tangent lines only where they intersect; so, the two graphs intersect at x^2 = Ax^2 + C





x^2 - Ax^2 = C





x^2(1 – A) = C





x = +/- sqrt [ C/(1- A)]








Thus, the points of tangency are





Point 1: x = sqrt [C/(1- A)] and y = C/(1- A)





Point 2: x = - sqrt [C/(1- A)] and y = C/(1- A)








The slope of the tangent line is then dy/dx = 2x





So slope m1 = 2 sqrt [ C/(1- A)] and slope m2 = - 2 sqrt [ C/(1- A)].





Let’s make things easier and call k = C/(1 – A)





Therefore the eqn of the tangent line at Point 1 is





y – k = 2sqrt(k)[ x – sqrt (k)]





y – k = 2sqrt(k)x – 2k





y = 2sqrt(k)x – k








The eqn of the tangent line at Point 2 is





y – k = - 2sqrt(k)[ x – sqrt (k)]





y – k = - 2sqrt(k)x + 2k





y = - 2sqrt(k)x + 3k





Now replace k and simplify.