Find the angle of intersection of the following curves :
`(i)y^2=xa n dx^2=y`
`(ii)y=x^2a n dx^2+y^2=20`
`(iii)2y^2=x^3a n dy^2=32 x`

Explanation

`(i)y^2=x and x^2=y`
first we have to find point of intersection
by equating both equations
`x^4=x`
`=>x(x^3-1)=0`
`=>x=0 or x=1`
at `x=0,y=0; x=1,y=1`
∴ point of intersection are (0,0) and (1,1)
Now,` y^2=x and x^2=y`
differentiate both the equation,
`(dy)/(dx)=1/(2y) and (dy)/(dx)=2x`
if we take `x=1 and y=1` then
`m_1=1/2 and m_2=2`
we know that `tan theta=abs((m_2-m_1)/(1+m_1*m_2))`
`=(3/2)/2`
`tan theta=3/4`
`theta=tan^-1(3/4)`


`(ii)y=x^2a n dx^2+y^2=20`
first we have to find point of intersection
by equating both equations
`=>y+y^2=12`
by factorization we get ,
`=>y=-5 or y=4`
at `y=-5 , x = sqrt(-5) and at y=4 , x = pm2`
differentiate both the equation,
`(dy)/(dx)=2x or (dy)/(dx)=-x/y`
`=>m_1=2x and m_2=-x/y`
if we take `x=2 and y=4` then
`m_1=4 and m_2=-1/2`
we know that `tan theta=abs((m_2-m_1)/(1+m_1*m_2))`
`=9/2`
`=>tan theta = 9/4`
`=>theta = tan^-1(9/4)`


`(iii)2y^2=x^3a n dy^2=32 x`
differentiate both the equation,
`(dy)/(dx)=(3x^2)/(4y) and (dy)/(dx)=32/(2y)`
`=>m_1=(3x^2)/(4y) and m_2=16/y`
From these given equation we get,
`2(32x)=x^3`
`=>x^3-64x=0`
`=>x=0 or x=pm8`
at `x=0 , y=0 ` and at `x=pm8, y=pm16`
when `x = 0, y = 0 `
`m_1=0 and m_2=oo`
we know that `tan theta=abs((m_2-m_1)/(1+m_1*m_2))`
`=oo`
`=>tan theta=00`
`=>theta = tan^-1(oo)`
`=>theta =pi/2`
similarly if we put `x=pm8, y=pm16`
then `theta=tan^-1(1/2)`