Geo5.2.24

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Find the locus of the point of intersection of the two tangents to the ellipse b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}\,,which include an angle theta.

Given the ellipse {\frac  {x^{2}}{a^{2}}}+{\frac  {y^{2}}{b^{2}}}=1\,

Let y=mx+{\sqrt  {a^{2}m^{2}+b^{2}}}\, be a tangent passing through a pointP(x_{1},y_{1})\,

Readjusting the variables,we have

m^{2}(x_{1}^{{2}}-a^{2})-2mx_{1}y_{1}+y_{1}^{{2}}-b^{2}\,

Let the roots of the quadratic equation be m_{1},m_{2}\,

Two tangents pass thro'P.If their inclinations are \alpha ,\beta \, then \tan \alpha =m_{1},\tan \beta =m_{2}\,

Given \theta =\alpha -\beta \,

\tan \theta =\tan(\alpha -\beta )\,

\tan \theta ={\frac  {\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }}={\frac  {m_{1}-m_{2}}{1+m_{1}m_{2}}}\,

Simplifying, we get

(1+m_{1}m_{2})^{2}\tan ^{2}\theta =(m_{1}-m_{2})^{2}=(m_{1}+m_{2})^{2}-4m_{1}m_{2}\,

[1+{\frac  {y_{1}^{{2}}-b^{2}}{x_{1}^{{2}}-a^{2}}}]\tan ^{2}\theta =[{\frac  {2x_{1}y_{1}}{x_{1}^{{2}}-a^{2}}}]^{2}-4({\frac  {y_{1}^{{2}}-b^{2}}{x_{1}^{{2}}-a^{2}}})\,

(x_{1}^{{2}}+y_{1}^{{2}}-a^{2}-b^{2})^{2}\tan ^{2}\theta =4x_{1}^{{2}}y_{1}^{{2}}-4(y_{1}^{{2}}-b^{2})(x_{1}^{{2}}-a^{2})\,

Therefore,the locus of P is (x^{2}+y^{2}-a^{2}-b^{2})^{2}\tan ^{2}\theta =4x^{2}y^{2}-4(y^{2}-b^{2})(x^{2}-a^{2})\,


Main Page:Geometry:The Ellipse