Answer to Question #125487 in Calculus for Ero

Question

Find all points on the graph of the equation x2-y2=2x+4y where the tangent line is horizontal. Does the graph have any vertical asymptote? Sketch the graph

Answer 1

x²-y²=2x+4y

x²-y²(x)-2x-4y(x)=0

=>

(x²-y²(x)-2x-4y(x))'=0

2x-2y(x)y'(x)-2-4y'(x)=0

The condition y'(x)=0 is true in points where the tangent line is horizontal.

If y'(x)=0 then 2x-2=0 => x=1

1²-y²-2*1-4y=0 =>

y²+4y+1=0

y₁=-2-3^1/2, y₂=-2+3^1/2

(1;-2-3^1/2), (1;-2+3^1/2) are points where the tangent line is horizontal.

(x-1)²-(y+2)²=-3

x-1=y+2 =>y=x-3

x-1=-(y+2) => y=-x-1

Asymptotes :

y=x-3, y=-x-1

So, the graph does not have any vertical asymptote.

Answer:

(1;-2-3^1/2), (1;-2+3^1/2) are points where the tangent line is horizontal;

The graph does not have any vertical asymptote