If $λ$ be a variable parameter, prove that the locus of the vertices of the hyperbolas given by the equation $x^{2} - y^{2} + λ x y = a^{2}$ is the curve $(x^{2} + y^{2})^{2} = a^{2} (x^{2} - y^{2}) .$

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Solution

The given equation is;-
$x^{2} - y^{2} + λ x y = a^{2}$
Here, $tan 2 θ = \frac{2 h}{a - b} = \frac{λ}{2} . . . . . .1$
and hence $r^{2} = \frac{a^{2} (1 + {tan}^{2} θ)}{1 - {tan}^{2} θ + λ tan θ}$
which is the polar equation of the locus of the vertex
or $r^{2} = \frac{a^{2}}{\frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ} + \frac{λ}{2} \frac{2 tan θ}{1 + {tan}^{2} θ}}$ and $\frac{λ}{2} = tan 2 θ$
or $r^{2} = \frac{a^{2}}{cos 2 θ + tan 2 θ sin 2 θ} = \frac{a^{2} cos 2 θ}{{cos}^{2} 2 θ + {sin}^{2} 2 θ}$
$= a^{2} cos 2 θ$
$= a^{2} ({cos}^{2} θ - {sin}^{2} θ)$
changing in to the cartesian co ordinates,
i.e., putting $x = r cos θ; y = r s i n θ$
and $x^{2} + y^{2} = r^{2}$ , we get
The required equation as
${(x^{2} + y^{2})}^{2} = a^{2} (x^{2} - y^{2})$

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