Question

If vertex of two parabolas `y=ax^2+4bx+c` and `y=-ax^2+4mx-c` are same, prove that `4b^2=ac`?

Answer 1

Please see below.

Explanation:

Let us convert the two equations into vertex form.

`y=ax^2+4bx+c`

= `a(x^2+4b/ax+(2b/a)^2-4b^2/a^2)+c`

= `a(x+(2b)/a)^2-4b^2/a+c`

= `a(x+(2b)/a)^2-(4b^2-ac)/a`

and its vertex is `(-(2b)/a,-(4b^2-ac)/a)`

`y=-ax^2+4mx-c`

= `-a(x^2-4m/ax+(2m/a)^2-4m^2/a^2)-c`

= `-a(x-(2m)/a)^2+4m^2/a-c`

= `-a(x-(2m)/a)^2+(4m^2-ac)/a`

and its vertex is `((2m)/a,(4m^2-ac)/a)`

As vertex of the two is same, we have

`-(2b)/a=(2m)/a` or `b=-m`

This means `b` amd `m` have opposite sign i.e. `bm<0`

or `b+m=0` and `b^2=m^2`

and `-(4b^2-ac)/a=(4m^2-ac)/a`

or `-(4b^2)/a+c=(4m^2)/a-c`

or `2c=4/a(m^2+b^2)`

but `b^2=m^2`, hence we have `2c=(8b^2)/a` or `ac=4b^2`

as `b^2` is always positive, `ac>0`

Further `(b-m)^2=(b-(-b))^2=4b^2=ac`