Geo5.3.31

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Tangents drawn from (\alpha ,\beta )\, to the hyperbola {\frac  {x^{2}}{a^{2}}}-{\frac  {y^{2}}{b^{2}}}=1\, make an angle \theta _{1},\theta _{2}\, with the x-axis.If \tan \theta _{1}\tan \theta _{2}=1\, prove that\alpha ^{2}-\beta ^{2}=a^{2}+b^{2}\,

Any equation of tangent to the given hyperbola is y=mx\pm {\sqrt  {a^{2}m^{2}-b^{2}}}\,

Let the given hyperbola pass thro' (\alpha ,\beta )\,

Then

\beta =m\alpha +{\sqrt  {a^{2}m^{2}-b^{2}}}\,

(\beta -m\alpha )^{2}=a^{2}m^{2}-b^{2}\,

(\alpha ^{2}-a^{2})m^{2}-2\alpha \beta m+(\beta ^{2}+b^{2})=0\,

If m1,m2 are the roots of the above quadratic, m_{1}m_{2}={\frac  {\beta ^{2}+b^{2}}{\alpha ^{2}-a^{2}}}\,

Given m_{1}=\tan \theta _{1},m_{2}=\tan \theta _{2},m_{1}m_{2}=1\,

1={\frac  {\beta ^{2}+b^{2}}{\alpha ^{2}-a^{2}}}\,

\alpha ^{2}-\beta ^{2}=a^{2}+b^{2}\,


Main Page:Geometry:Hyperbola