Geo5.3.49

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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\,</math>

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}\,


Main Page:Geometry:Hyperbola