Geo5.3.29

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Show that the locus of the foot of perpendicular from the centre of the hyperbola {\frac  {x_{2}}{a^{2}}}-{\frac  {y^{2}}{b^{2}}}=1\, on a variable tangent is (x^{2}+y^{2})^{2}=a^{2}x^{2}-b^{2}y^{2}\,

Any tangent to the given hyperbola is y=mx\pm {\sqrt  {a^{2}m^{2}-b^{2}}}\,

The equation to the perpendicular to the above and passing thro'the origin is my+x=0,m={\frac  {-x}{y}}\,

The required locus is obtained by eliminating m in the above two equations,

y=x(-{\frac  {x}{y}})\pm {\sqrt  {a^{2}({\frac  {x^{2}}{y^{2}}}-b^{2})}}\,

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

Squaring on both sides,we get

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

This is the required locus.


Main Page:Geometry:Hyperbola