SOLUTION: find equation of normal to hyperbola x2-3y2=144 at positive end of latus rectum

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Question 951534: find equation of normal to hyperbola x2-3y2=144 at positive end of latus rectum
Answer by MathLover1(20850) About Me  (Show Source):
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x%5E2-3y%5E2=144+
x%5E2%2F144-3y%5E2%2F144=144+%2F144
x%5E2%2F144-y%5E2%2F48=1
center: (0,0)
a%5E2=144 => a=sqrt%28144%29=12
b%5E2=48 => b=sqrt%2848%29=6.93
then c%5E2=a%5E2%2Bb%5E2 =>c%5E2=144%2B48=192 => c=sqrt%28192%29
=> c=sqrt%2864%2A3%29=> c=8sqrt%283%29
The Latus Rectum is the line through the focus and parallel to the directrix.
The length of the Latus Rectum is:
2b%5E2%2Fa=%282%2A48%29%2F12=48%2F6=8
positive end of latus rectum: (L,4)


positive end of latus rectum at Point (L,4):
(c+ ,b%5E2%2Fa)=(8sqrt%283%29 ,4)
normal at (8sqrt%283%29 ,4)=(x%5B1%5D,y%5B1%5D) is:
a%5E2x%2Fx%5B1%5D%2Bb%5E2y%2Fy%5B1%5D=a%5E2%2Bb%5E2
144x%2F8sqrt%283%29%2B48y%2F4=144%2B48
6sqrt%283%29x%2B12y+=+192 .......divide by 6
sqrt%283%29x%2B2y-32+=0