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

Options

A. four distinct points is `agtbgt1`
B. no point if `blt-1`
C. two distinct points if `-1ltblt1`
D. one point if `b=1`

Answer

A, B, C

Explanation

`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