Question

x²+y²=2x,y²=x,Find the measure of the angle between curves,if they intersect.

Answer 1

x² + y² = 2x , y² = x

first of all find intersecting points of given curves.
x² + y² = 2x , y² = x

x² + x = 2x
x² = x => x = 0, 1
put x = 0, in equation y² = x => y = 0,
put x = 1 , in equation y² =x => y = ±1

so, points are (0,0) , (1, -1) and (1 ,1)

now, find slopes of tangent of curve ,
x² + y² = 2x
differentiate with respect to x,
2x + 2y.dy/dx = 2
x + y.dy/dx = 1
dy/dx = (1 - x)/y

at (0,0) $m_1=\frac{dy}{dx}|_{(0,0)}$ =∞
at (1, -1) $m_1'=\frac{dy}{dx}_{(1,-1)}$=0
at (1,1) $m_1"=\frac{dy}{dx}_{(1,1)}$=0

again, y² = x
differentiate with respect to x,
2y.dy/dx = 1
dy/dx = 1/2y
at (0,0) $m_2=\frac{dy}{dx}_{(0,0)}$=∞
at (1,-1) $m_2'=\frac{dy}{dx}_{(1,-1)}$=-1/2
at (1,1) $m_2"=\frac{dy}{dx}_{(1,1)}$ = 1/2

now, angle better curves at (0,0)
= $tan^{-1}\frac{|\infty-\infty|}{1+\infty.\infty}$
= 0°

angle between curves at (1, -1)
= $tan^{-1}\frac{|-1/2-0|}{1+0.-1/2}$
= $tan^{-1}\frac{1}{2}$

angle between curves at (1,1)
= $tan^{-1}\frac{|1/2-0|}{1+0.1/2}$
= $tan^{-1}\frac{1}{2}$

hence, angle between curves : 0° and tan^-1(1/2)

Answer 2

Dear Student:

Clearly (0,0) is one point.

And angle is 0 degree.

Because slope of first curve is ∞

and second curve is ∞

Use formula of tangent angle.

See the attachment.