The angle between two curves at the point of intersection is the angle b/w their tangents:

Tan (theta) =m

Y

x=y

Put in x

Y

By solving we get;

y=0 and y=4a

When y=0 then x=0 we get (0, 0)

When y=4a then x=4a we get (4a, 4a)

Y

Differentiate it

2ydy/dx=4a

dy/dx=2a/y

at point (4a, 4a)

m

x

y=1/4a x

dy/dx= x/

at the point (4a, 4a)

m

let theta be the angle b/w them

tan(theta)=m

tan(theta)=3/4

theta=tan-1(3/2)

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Tan (theta) =m

_{2}-m_{1}/ 1-m_{1}m_{2}where m_{1}and m_{2}are the slopesY

^{2}=4ax x^{2}=4ayx=y

^{2}/4axPut in x

^{2}=4ayY

^{4}/16a^{2 }=4ayBy solving we get;

y=0 and y=4a

When y=0 then x=0 we get (0, 0)

When y=4a then x=4a we get (4a, 4a)

Y

^{2}=4axDifferentiate it

2ydy/dx=4a

dy/dx=2a/y

at point (4a, 4a)

m

_{1}=2a/4a=1/2x

^{2}=4ayy=1/4a x

^{2}dy/dx= x/

^{2}aat the point (4a, 4a)

m

_{2}=4a/2a= 2let theta be the angle b/w them

tan(theta)=m

_{2}-m_{1}/1-m_{1}m_{2}tan(theta)=3/4

theta=tan-1(3/2)

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