Geo4.36

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Prove that the locus of a point tangents from which to the circle x^{2}+y^{2}=a^{2}\, are inclined at an angle alpha is (x^{2}+y^{2}-2a^{2})^{2}(\tan \alpha )^{2}=4a^{2}(x^{2}+y^{2}-a^{2})\,.

Let Px1,y1) be any point on locus.

Hence

\tan \alpha ={\frac  {2a({\sqrt  {x_{1}^{{2}}+y_{1}^{{2}}-a^{2}}})}{{\sqrt  {x_{1}^{{2}}+y_{1}^{{2}}-2a^{2}}}}}\,

(x_{1}^{{2}}+y_{1}^{{2}}-2a^{2})\tan \alpha =2a({\sqrt  {x_{1}^{{2}}+y_{1}^{{2}}-a^{2}}})\,

Squaring on both sides,we get

(x_{1}^{{2}}+y_{1}^{{2}}-2a^{2})^{2}(\tan ^{{2}}\alpha )=4a^{2}(x_{1}^{{2}}+y_{1}^{{2}}-a^{2})\,

Hence the required.


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