Focuses of the hyperbola F 1 ( 4 , 2 ) , F 2 ( − 1 , − 10 ) and the tangent equation 3 x + 4 y − 5 = 0 are given. Find the semiaxes of the hyperbola.I’ve tried using the definition of hyperbola and the equation for the tangent line to the quadratic curve to form some sort of system of equations, but it didn’t work out.

Question

Focuses of the hyperbola       F    1    (  4  ,     2  )  ,         F    2    (  −  1  ,     −  10  ) and the tangent equation   3  x  +  4  y  −  5  =  0 are given. Find the semiaxes of the hyperbola.I’ve tried using the definition of hyperbola and the equation for the tangent line to the quadratic curve to form some sort of system of equations, but it didn’t work out.

Nia Molina · Accepted Answer

Step 1      (    x    −    4          )      2        +    (    y    −    2          )      2        −      (    x    +    1          )      2        +    (    y    +    10          )      2        =  ±  2  a      (    x    −    4          )      2        +    (    y    −    2          )      2        =  ±  2  a  +      (    x    +    1          )      2        +    (    y    +    10          )      2        (  x  −  4      )    2    +  (  y  −  2      )    2    =  4      a    2    ±  4  a      (    x    +    1          )      2        +    (    y    +    10          )      2        +  (  x  +  1      )    2    +  (  y  +  10      )    2    (  (  x  −  4      )    2    +  (  y  −  2      )    2    −  4      a    2    −  (  (  x  +  1      )    2    +  (  y  +  10      )    2    )      )    2    =  16      a    2    (  (  x  +  1      )    2    +  (  y  +  10      )    2    )  (  −  4      a    2    −  10  x  −  24  y  −  81      )    2    =  16      a    2    (  (  x  +  1      )    2    +  (  y  +  10      )    2    )  16      a    4    −  16      a    2        x    2    +  48      a    2    x  −  16      a    2        y    2    −  128      a    2    y  −  968      a    2    +  100      x    2    +  480  x  y  +  1620  x  +  576      y    2    +  3888  y  +  6561  =  0Dual:  −  16      a    2    (  2  a  −  13  )  (  2  a  +  13  )  (  4      a    2        y    2    −  89      y    2    +  168  x  y  +  32  y  +  4      a    2        x    2    −  153      x    2    −  12  x  −  4  )  =  0The tangent is   −      3    5    x  −      4    5    y  +  1  =  0 so setting   (  −      3    5    ,  −      4    5    ) into the dual we get       a    2    =      269    20    .That is the equation is  2255      y    2    +  3000  x  y  +  13540  y  −  720      x    2    +  14160  x  −  22276  =  0or  2255  (  y  +  4      )    2    +  3000  (  x  −      3    2    )  (  y  +  4  )  −  720  (  x  −      3    2        )    2    −  38736  =  0The matrix has eigenvalues 2880, -1345, so the canonical form is  (  20      y          ′      2        )      /    269  −  (  5      x          ′      2        )      /    144  =  1  ,so the semi-axes are             269      20        ,      12          5        .

juanberrio8a · Accepted Answer

Step 1The product of distances from focis to any tangent line is the square of conjugate semiaxis.The distances from       F    1   and       F    2   to the tangent line are respectfully   15      /    5 and   48      /    5 , so   b  =  12      /        5   .The distance from       F    1   to       F    2   is equal to 2c, then   4      c    2    =      13    2   .Then   a  =            c      2        −          b      2        =      269        /    (  2      5    ) .

Focuses of the hyperbola F 1 </msub> ( 4 , <mtext> </mtext> 2

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