SOLUTION: Halle la ecuación canónica de la hipérbola que tiene sus Focos en (±3,0) y lado recto |𝐿R| = 5

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Question 1194079: Halle la ecuación canónica de la hipérbola que tiene sus Focos en (±3,0) y lado recto |𝐿R| = 5
Answer by MathLover1(20850) About Me  (Show Source):
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Find the canonical equation of the hyperbola that has its Foci at (±3,0) and straight side |SS| = 5
given:
Foci at (±3,0) =>c=3
=> Since, foci is on x-axis, equation of hyperbola is of the form
x%5E2%2Fa%5E2+-+y%5E2%2Fb%5E2=1+
we know that c%5E2=a%5E2%2Bb%5E2
so far we have 3%5E2=a%5E2%2Bb%5E2=>9=a%5E2%2Bb%5E2
if given straight side |SS| =+5, means the length of the latus rectum of the hyperbola is 5
so, %282b%5E2%29%2Fa=5.........solve for a
%282b%5E2%29=5a
a=%282b%5E2%29%2F5

go to =>9=a%5E2%2Bb%5E2, substitute a
9=%28%282b%5E2%29%2F5%29%5E2%2Bb%5E2.......solve for b
9=%284b%5E4%29%2F25%2Bb%5E2
9%2A25=4b%5E4%2B25b%5E2
4b%5E4%2B25b%5E2-225=0
%28b%5E2+-+5%29+%284+b%5E2+%2B+45%29+=+0
solutions:
=>b%5E2+=5
or
b%5E2=+-45%2F5
Since, a is distance, it cant be negative.
using positive solution. find a
a=%282%2A5%29%2F5
a=2=>a%5E2=4

and your equation is:

x%5E2%2F4+-+y%5E2%2F5=1+