Find the angle of intersection of the curves y`=4-x^(2)` and `y=x^(2)`

Options

A. `theta = tan^(-1)((4sqrt(2))/(7))`
B. `theta = tan^(-1)((sqrt(2))/(7))`
C. `theta = tan^(-1)(4sqrt(2))`
D. `theta = tan^(-1)(sqrt(2))`

Answer

A

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` and ` m_(2)=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 for point `(sqrt(2), 2)`
`m_(1)=-2x=-2sqrt(2)` , `m_(2)=2x=2sqrt(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))`
And for point`(-sqrt(2),2)`.
`m_(1) = -(2x).(-sqrt(2))= 2sqrt(2)`
and `m_(2)=2x=-2sqrt(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))`