Focuses of the hyperbola F 1 </msub> ( 4 , <mtext>&#xA0;</mtext> 2

anginih86

anginih86

Answered question

2022-06-14

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.

Answer & Explanation

Nia Molina

Nia Molina

Beginner2022-06-15Added 21 answers

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 = 0
Dual:
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 ) = 0
The 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 = 0
or
2255 ( y + 4 ) 2 + 3000 ( x 3 2 ) ( y + 4 ) 720 ( x 3 2 ) 2 38736 = 0
The 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

juanberrio8a

Beginner2022-06-16Added 7 answers

Step 1
The 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 ) .

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