Geo5.3.26

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Find the line passing through the point (-2,1) and conjugate to the line 8x+3y-4=0\, w.r.t 2x^{2}-y^{2}=1\,

Given hyperbola {\frac  {x^{2}}{2}}-{\frac  {y^{2}}{4}}=1\,

Let P(x_{1},y_{1})\, be the pole of the line 8x+3y-4=0\,

But the equation to polar of P w.r.t the hyperbola is 2xx_{1}-yy_{1}=4\,

Now the above two equations represent the same line

{\frac  {2x_{1}}{8}}={\frac  {-y_{1}}{3}}={\frac  {-4}{-4}}\,

x_{1}=4,y_{1}=-3\,

(4,-3)\,

Therefore,the equation to the line joining P(4,-3) with (-2,1) is y+3={\frac  {4}{-6}}(x-4)\,

2x+3y+1=0\,

The required conjugate line is 2x+3y+1=0\,


Main Page:Geometry:Hyperbola