Geo5.3.49

Show that the locus of the foot of the perpendicular drawn from the centre of the hyperbola ${\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}=1\,$ on any normal to it is $(a^{2}y^{2}-b^{2}x^{2})(x^{2}+y^{2})^{2}=(a^{2}+b^{2})x^{2}y^{2}\,$

Equation to the normal at theta is \frac{ax}{\sec\theta}+\frac{by}{\tan\theta}=a^2+b^2\,[/itex]

$ax\cos \theta +by\cot \theta =a^{2}+b^{2}\,$

Equation to the line passing thro'origin and perpendicular to the above equation is

${\frac {bx}{\tan \theta }}-{\frac {ay}{\sec \theta }}=0\,$

$\sin \theta ={\frac {bx}{ay}}\,$

$\cos \theta ={\frac {{\sqrt {a^{2}y^{2}-b^{2}x^{2}}}}{ay}}\,$

$\cot \theta ={\frac {{\sqrt {a^{2}y^{2}-b^{2}x^{2}}}}{bx}}\,$

The locus of the foot of the perpendicular is obtained by eliminating theta between the above and the normal equation,

$[{\frac {x}{y}}+{\frac {y}{x}}]({\sqrt {a^{2}y^{2}-b^{2}x^{2}}})=(a^{2}+b^{2})\,$

$(x^{2}+y^{2})^{2}(a^{2}y^{2}-b^{2}x^{2})=(a^{2}+b^{2})x^{2}y^{2}\,$