Question

Find the angle of intersection of the following pair of curve:
y = x + 1,
x² - 2xy + y² = x

Answer 1

\[y=x+1\quad,\quad x^2-2xy+y^2=x\]\[x^2-2xy+y^2=x\]\[(x-y)^2=x\]\[(-1)^2=x=1\quad,\quad y=2\]\[(x^2-2xy+y^2)'=(x)'\]\[2x-2y-2xy'+2yy'=1\]\[2(x-y)-2(x-y)y'=1\]\[1-y'=\frac{1}{2(x-y)}\]\[y'=1-\frac{1}{2(x-y)}=1+\frac{1}{2(y-x)}=\frac{3}{2}\]\[\tan\alpha=\frac{1.5-1}{1+1.5}=0.2\quad,\quad\alpha=\arctan{0.2}=11.3^{\circ}\]

Answer 2

Thank you! :)