The angle between two curves at the point of intersection is the angle b/w their tangents:
Tan (theta) =m2-m1/ 1-m1m2 where m1 and m2 are the slopes
Y2=4ax x2=4ay
x=y2/4ax
Put in x2=4ay
Y4/16a2 =4ay
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)
Y2=4ax
Differentiate it
2ydy/dx=4a
dy/dx=2a/y
at point (4a, 4a)
m1=2a/4a=1/2
x2=4ay
y=1/4a x2
dy/dx= x/2a
at the point (4a, 4a)
m2=4a/2a= 2
let theta be the angle b/w them
tan(theta)=m2-m1/1-m1m2
tan(theta)=3/4
theta=tan-1(3/2)
www.regentsprep.org
Tan (theta) =m2-m1/ 1-m1m2 where m1 and m2 are the slopes
Y2=4ax x2=4ay
x=y2/4ax
Put in x2=4ay
Y4/16a2 =4ay
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)
Y2=4ax
Differentiate it
2ydy/dx=4a
dy/dx=2a/y
at point (4a, 4a)
m1=2a/4a=1/2
x2=4ay
y=1/4a x2
dy/dx= x/2a
at the point (4a, 4a)
m2=4a/2a= 2
let theta be the angle b/w them
tan(theta)=m2-m1/1-m1m2
tan(theta)=3/4
theta=tan-1(3/2)
www.regentsprep.org