Geo5.2.37

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Find the equation of the pair of tangents to the ellipse x^{2}+3y^{2}=3\, from the point (2,-1)\,.

Given ellipse is S\equiv x^{2}+3y^{2}-3=0,(x_{1},y_{1})=(2,-1)\,

S_{1}\equiv x(2)+3y(-1)-3=2x-3y-3\,

S_{{11}}\equiv (2)^{2}+3(-1)^{2}-3=4\,

Therefore equation to the pair of tangents is S_{1}^{{2}}=SS_{{11}}\,

(2x-3y-3)^{2}=(x^{2}+3y^{2}-3)4\,

4x^{2}+9y^{2}-12xy+18y-12x=4x^{2}+12y^{2}-12\,

y^{2}+4xy+4x-6y-7=0\,


Main Page:Geometry:The Ellipse