Asked by John
what is the length of the latus rectum of x²-4y²-8x+64y-256=0
Answers
Answered by
mathhelper
x²-4y²-8x+64y-256=0
x^2 - 8x + 16 - 4(y^2 - 16y + 64) = 256 + 16 - 256
(x - 4)^2 - 4(y - 8)^2 = 16
(x-4)^2 / 16 - (y - 8)^2 / 4 = 1
a = 4, b = 2
the latus rectum is the length of the vertical "chord" passing through the
focal points. That length is 2b^2 / a
= 2(4)/4 = 2 units long
x^2 - 8x + 16 - 4(y^2 - 16y + 64) = 256 + 16 - 256
(x - 4)^2 - 4(y - 8)^2 = 16
(x-4)^2 / 16 - (y - 8)^2 / 4 = 1
a = 4, b = 2
the latus rectum is the length of the vertical "chord" passing through the
focal points. That length is 2b^2 / a
= 2(4)/4 = 2 units long
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