Geo5.3.27

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Find the equation to the line passing through (1,2) and conjugate to the line 3x+2y+6=0\, w.r.t hyperbola 3x^{2}-4y^{2}=12\,

Let P be the pole of the line 3x+2y+6=0\,

But the equation to polar of P w.r.t given hyperbola is 3xx_{1}-4yy_{1}=12\,

The above two represent the same line,

{\frac  {3x_{1}}{3}}={\frac  {-4y_{1}}{2}}={\frac  {-12}{6}}\,

x_{1}=-2,y_{1}=1\,

The equation to the line joining (1,2),(-2,1)\, is

y-2={\frac  {1-2}{-2-1}}(x-1)\,

-3y+6=-x+1\,

x-3y+5=0\,

Therefore,the required conjugate line is x-3y+5=0\,


Main Page:Geometry:Hyperbola