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Show that the poles of the tangents of y^{2}=4kx(k>0)\, w.r.t the ellipse {\frac  {x^{2}}{a^{2}}}+{\frac  {y^{2}}{b^{2}}}=1\, lie on a parabola.

Let P(x_{1},y_{1})\, be the pole of a tangent l=0\, of y^{2}=4kx\, w.r.t the ellipse S=0\, where l=y-(mx+{\frac  {k}{m}})\,

Then the polar of P w.r.t S=0 is {\frac  {xx_{1}}{a^{2}}}+{\frac  {yy_{1}}{b^{2}}}-1=0\,

Since S_{1}=0,l=0\, represents the same line,comparing coefficients we get

{\frac  {x_{1}}{a^{2}m}}={\frac  {y_{1}}{-b^{2}}}={\frac  {-m}{k}}\,

{\frac  {x_{1}}{a^{2}m}}={\frac  {y_{1}}{-b^{2}}},m={\frac  {-b^{2}x_{1}}{a^{2}y_{1}}}\,

{\frac  {y_{1}}{-b^{2}}}={\frac  {-m}{k}},m={\frac  {ky_{1}}{b^{2}}}\,

{\frac  {-b^{2}x_{1}}{a^{2}y_{1}}}={\frac  {ky_{1}}{b^{2}}}\,

y_{1}^{{2}}={\frac  {-b^{4}x_{1}}{ka^{2}}}\,

Hence the pole P(x_{1},y_{1})\, lies on the parabola y^{2}={\frac  {-b^{4}}{ka^{2}}}(x_{1})\,

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