Question

Show that if B>0, then the graph of x²+Bxy=F, is a hyperbola if F≠0, and two intersecting lines if F=0.

Answer 1

A conic in the two-dimensional plane is the solution set of an equation of the form

$$ax^2 + bxy + cy^2 + dx + ey +f = 0$$

for real parameters $a,b,c,d,e$ and $f$, where at least one of $a,b$ and $c$ is not zero.

If $b^2-4ac>0,$ then the graph is a hyperbola (or two intersecting lines).

We have

$$x^2+Bxy=F$$

$$a=1, b=B, c=0, f=-F$$

$$b^2-4ac=B^2-4(1)(0)=B^2>0, B\not=0$$

Therefore the graph of $x^2+Bxy=F,$ is a hyperbola or two intersecting lines., if $B\not=0.$

If $F\not=0,$ we have a hyperbola

$$y=-\dfrac{x}{B}+\dfrac{F/B}{x}$$

If $F=0,x^2+Bxy=0,$ we have two intersecting lines

$$x=0\ or\ y=-\dfrac{1}{B}x$$