Find the angle of intersection of curve `y=4-x^2` and `y=x^2`

Explanation

We have, `y=4-x^(2)`…………..(i)
and `y=x^(2)`……………….(ii)
`rArr (dy)/(dx) = -2x`
and `(dy)/(dx) = 2x`
`rArr m_(1)=-2x`
From Eqs. (i) and (ii), `x^(2)=4-x^(2)`
`rArr 2x^(2)=4`
`rArr x^(2)=2`
`therefore x=+-sqrt(2)`
`therefore y=x^(2)=(+-sqrt(2))^(2)=2`
So, the points of intersection are `(sqrt(2),2)` and `(-sqrt(2),2)`.
and `m_(2)=2x=2sqrt(2)`
and for points `(sqrt(2), 2)` and `(-sqrt(2),2)`.
For point `(+sqrt(2),2)` `m_(1) = -2x.sqrt(2)=-2sqrt(2)`
and `m_(2)=2x=2sqrt(2)`
and for point `(sqrt(2),2)`, `tantheta= |(m_(1)-m_(2))/(1+m_(1)m_(2))|=|(-2sqrt(2)-2sqrt(2))/(1-2sqrt(2).2sqrt(2))| = |(-4sqrt(2))/(-7)|`
`theta = tan^(-)(4sqrt(2))/(7)`