Geo5.3.16

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Prove that the line 21x+5y-116=0\, touches the hyperbola 7x^{2}-5y^{2}=232\, and find the point of contact.

From the Given hyperbola a^{2}={\frac  {232}{7}},b^{2}={\frac  {232}{5}}\,

From the given line l=21,m=5,n=116\,

The condition of tangency is a^{2}l^{2}-b^{2}m^{2}=n^{2}\,

{\frac  {232}{7}}(21)^{2}-{\frac  {232}{5}}(25)=116^{2}\,

232(63-5)=116^{2}\,

116^{2}=116^{2}\,

Therefore,the line touches the hyperbola.

Point of contact is \left(-{\frac  {(232)(21)}{(7)(116)}},{\frac  {(232)(5)}{(5)(116)}}\right)\,

\left(6,2\right)\,


Main Page:Geometry:Hyperbola