Question

Find the angle of intersection of the curves $xy=a^2$ and $x^2+y^2=2a^2$.

Answer 1

$xy=a^{2}y=\frac{a^2}{x}{x}^2+y^2=2a^2{x}^2+\frac{a^4}{x^2}=2a^2{x}^4-2a^2{x}^2+a^4=0{(x^2-a^2)}^2=0\therefore x^2=a^{2}x=\pm a\therefore y=\pm ay+x{y}^{\prime }=0y^{\prime }=-\frac{y}{x}=-1atx=\pm a,y=\pm a{x}^2+y^2=2a^{2}2x+2y{y}^{\prime }=0y^{\prime }=-\frac{x}{y}=-1atx=\pm a,y=\pm a$

At x=$\pm$a, y=$\pm$a, slopes are equal.

$\therefore$ Angle of intersection of the curves = 0 [$\because$ Slopes are equal]