Answer to Question #125671 in Analytic Geometry for Samuel kassapa

Question

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

Answer 1

x²-y²=2x+4y

y²+4y=x²-2x

y²+4y+4-4=x²-2x+1-1

(y+2)² =(x-1)²+3 conjugate hyperbole

two branches

y=((x-1)²+3)^1/2-2

y=-((x-1)²+3)^1/2-2

find the derivatives

y`=(x-1)/((x-1)²+3)^1/2

y`=-(x-1)/((x-1)²+3)^1/2

the derivative is 0 for

x=1 y₁=(3)^1/2-2 y₂=-(3)^1/2-2

The line x = x₀ cannot be a vertical asymptote if the function is continuous at the point x = x₀. Therefore, vertical asymptotes should be sought at the points of discontinuity of the function.