Geo5.3.2

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Find the equation to the hyperbola whose focus is (1,2)\, eccentricity {\sqrt  {3}}\, and directrix is 2x+y-1=0\,

Given S=(2,1),l=2x+y-1=0,e={\sqrt  {3}}\,

Equation to hyperbola is (x-2)^{2}+(y-1)^{2}=({\sqrt  {3}})^{2}[{\frac  {2x+y-1}{{\sqrt  {5}}}}]^{2}\,

(x^{2}+y^{2}-4x-2y+5)={\frac  {3}{5}}[2x+y-1]^{2}\,

5(x^{2}+y^{2}-4x-2y+5)=3(4x^{2}+y^{2}+1+4xy-2y-4x)\,

5x^{2}-12x^{2}+5y^{2}-3y^{2}-20x+12x-10y+6y-12xy+25-3=0\,

-7x^{2}+2y^{2}-8x-4y-12xy+22=0\,

7x^{2}-12xy-8x-4y+22=0\,


Main Page:Geometry:Hyperbola