SOLUTION: find equation of hyperbola where foci is (9,0) and hyperbola passes through (12,3)

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Question 928231: find equation of hyperbola where foci is (9,0) and hyperbola passes through (12,3)
Answer by MathLover1(20850) About Me  (Show Source):
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find equation of hyperbola where foci is (9,0) and hyperbola passes through (12,3)


The general formula of a hyperbola is:
x%5E2+%2F+a%5E2+-+y%5E2+%2F+b%5E2=+1

Since, (12, 3) is a point on the hyperbola, we have:

12%5E2+%2F+a%5E2+-+3%5E2+%2F+b%5E2+=+1
144+%2F+a%5E2+-+9+%2F+b%5E2+=+1........eq.(1)

Since the foci is (9,0), we have:

a%5E2+%2B+b%5E2+=+9%5E2

b%5E2+=+81+-+a%5E2.....eq.(2)

Substituting (2) into (1), we have:

144+%2F+a%5E2+-+9+%2F+%2881-a%5E2%29+=+1 ....common denominator is a%5E2%2881-a%5E2%29

%28144%2881-a%5E2%29-9a%5E2%29%2F%28a%5E2%2881-a%5E2%29%29=1

%28144%2881-a%5E2%29-9a%5E2%29=1%28a%5E2%2881-a%5E2%29%29

11664-144a%5E2-9a%5E2=81a%5E2-a%5E4

0=144a%5E2%2B81a%5E2%2B9a%5E2-a%5E4-11664

0=234a%5E2-a%5E4-11664

a%5E4-234a%5E2%2B11664=0....write -234a%5E2 as -162a%5E2-72x%5E2

a%5E4-162a%5E2-72x%5E2%2B11664=0

%28a%5E4-162a%5E2%29-%2872x%5E2%2B11664%29=0

a%5E2%28a%5E2-162%29-72%28x%5E2%2B162%29=0

%28a%5E2-162%29%28a%5E2-72%29+=+0
solutions:
=>%28a%5E2-162%29+=+0=>a%5E2+=+162
=>%28a%5E2-72%29+=+0=>a%5E2+=+72

Suppose a%5E2+=+162, from (2), we have:

b%5E2+=+81+-+162+=+-81=>rejected, since a square cannot be negative

Suppose a%5E2+=+72, from (2), we have:
b%5E2+=+81-72+=+9=> this solution is accepted

Thus, we have:

x%5E2%2F72+-y%5E2%2F9+=+1
foci is (9,0) and (-9,0)
hyperbola passes through (12,3)