Answer to Question #281150 in Calculus for Sajid

Question

The graph of(x²+y²)²=4(x²-y²) shown in the figure is called lemnicate find the points on the graph that correspond to x=1 find the equation of the tangent line to the graph at each point found in point a find the points on the graph at which the tangent is horizontal

Answer 1

at x=1:

$(1+y²)²=4(1-y²)$

$1+2y^2+y^4=4-4y^2$

$y^4+6y^2-3=0$

$y^2=\frac{-6+\sqrt{36+12}}{2}=-3+2\sqrt 3=0.46$

$y=\pm 0.68$

$2(x²+y²)(2x+2yy')=4(2x-2yy')$

$y'=\frac{4x-2x(x^2+y^2)}{2y(2+x^2+y^2)}$

equation of the tangent line:

$y-y_0=f'(x_0)(x-x_0)$

at point $(1,0.68)$ :

$y'=0.23$

$y-0.68=0.23(x-1)$

$y=0.23x+0.45$

at point $(1,-0.68)$ :

$y'=-0.23$

$y+0.68=-0.23(x-1)$

$y=-0.23x-0.45$

tangent is horizontal where y' = 0 :

$\frac{4x-2x(x^2+y^2)}{2y(2+x^2+y^2)}=0$

$x^2+y^2=2$ , then:

$4=4(2-2y^2)$

$y=\pm 1/\sqrt 2=\pm 0.70$

$x=\pm \sqrt {3/2}=\pm 1.22$

points: $(-1.22,0.7),(-1.22,-0.7),(1.22,0.7),(1.22,-0.7)$