Show that one can solve x3 d bx2 by intersecting the hyper

Show that one can solve x3 + d = bx2 by intersecting the hyperbolaxy=dandtheparabolay2+dxdb=0.As- suming that 3 d < b, determine the conditions on b and d that give zero, one, or two intersections of these two conics.

Solution

solution-

two conics xy = d and y2 + dx db = 0

Simply substitute y = d/x
in the other equation. we get

(d/x)^2+ dx db = 0 we will get

d^2/x^2 +dx-db=0 since which for the case x > 0 and d 0

we will multiply both sides by x^2/d   we will get

d^2/x^2 *x^2/d +dx*x^2/d-db*x/2/d=0

d+x^3-bx^2=0

we get     x^3+d=bx^2

To determine the condition on d and b for the conics to intersect in zero, one
or two points, one way is as follows

Using
the second derivative, it becomes clear that if b, d > 0, then x = 2b/3
gives the local minimum
and x = 0 gives the local maximum (since f(2b/3)= 2b > 0 and f(0) = 2b < 0) of the cubic