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Consider the circle x^(2)+y^(2)=1 and thhe parabola y=ax^(2)b(agt0). This circle and parabola intersect at

Answer»

FOUR distinct POINTS is `agtbgt1`
no POINT if `blt-1`
two distinct points if `-1ltblt1`
one point if `b=1`

SOLUTION :`x^(2)+(ax^(2)-b)^(2)=1`
`impliesa^(2)x^(4)+(1-2ab)x^(2)+(b^(2)-1)=0`
`impliesa^(2)t^(2)+(1-2ab)t+(b^(2)-1)=0`
`impliesf(t)=0`
`D=4a^(2)-4ab+1`
`agtbgt1impliesDgt0`, `f(0)gt0`
and `(2ab-1)/(2a^(2))gt0impliest_(1)gt0,t_(2)gt0`
`implies` four distinct real values of x
`blt-1impliesDgt0,f(0)gt0` and `(2ab-1)/(2a^(2))lt0`
`impliest_(1)lt0,t_(2)lt0implies` no real value of x
`-1ltblt1impliesf(0)lt0impliest_(1)gt0,t_(2)lt0`
`implies` two distinct real values of x


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