Question 1143176
 Find the values of a for which the curve {{{y= x^2}}} never touches the curve {{{y= a-(x-a)^2 }}}

{{{y= x^2 }}}
{{{y= a-(x-a)^2}}}
------------------------

{{{x^2 =a-(x-a)^2}}}
{{{x^2 =a-(x^2-2xa+a^2)}}}
{{{x^2 +(x^2-2xa+a^2)-a=0}}}
{{{x^2 +x^2-2xa+a^2-a=0}}}
{{{2x^2-2ax+a^2-a=0}}}

use discriminant {{{D=b^2-4ac }}}  
When {{{a}}},{{{ b}}}, and {{{c}}} are real numbers, {{{a <> 0}}} and discriminant is {{{negative}}}, then the roots α and β of the quadratic equation {{{ax^2 + bx + c = 0 }}}are unequal and {{{not}}} real. In this case, we say that the roots are {{{imaginary}}}.

in your case discriminant {{{D}}} is:

{{{2x^2-2ax+a^2-a=0}}}

since {{{a=2}}},{{{ b=-2a}}}, and {{{c=a^2-a}}}

{{{D=(-2a)^2-4*2(a^2-a)}}}

{{{D=4a^2-8(a^2-a)}}}

{{{D=4a^2-8a^2+8a}}}

{{{D=-4a^2+8a}}}

{{{D=-4a(a-2)}}}


if {{{D<0}}} => {{{-4a(a-2)<0}}} 

solutions:

if {{{-4a >0}}} =>{{{a<0}}}
if {{{(a-2)>0}}}=>{{{a>2}}}


so, your solution is: 
{{{a}}} is in interval ({{{-infinity}}}, {{{0}}}) 
or
 {{{a}}} is in interval ({{{2}}},{{{infinity}}})



check some values:

let's {{{a=-1}}} in interval ({{{-infinity}}}, {{{0}}}) 

{{{y= x^2 }}}
{{{y= -1-(x+1)^2}}}

graph:

{{{ graph( 600, 600, -10,10, -10, 10, x^2,-1-(x+1)^2) }}}

so, the curve {{{y= x^2}}} never touches the curve {{{y= a-(x-a)^2}}}


let's {{{a=3}}} in interval ({{{2}}},{{{infinity}}})

{{{y= x^2 }}}
{{{y= 3-(x-3)^2}}}

graph:

{{{ graph( 600, 600, -10,10, -10, 10, x^2,3-(x-3)^2) }}}

so, the curve {{{y= x^2}}} never touches the curve {{{y= a-(x-a)^2}}}