Question 1194079
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^2/a^2 - y^2/b^2=1 }}}

we know that {{{c^2=a^2+b^2}}}

so far we have {{{3^2=a^2+b^2}}}=>{{{9=a^2+b^2}}}

if given   straight side |SS| ={{{ 5}}}, means the length of the latus rectum of the hyperbola is {{{5}}}

so, {{{(2b^2)/a=5}}}.........solve for {{{a}}}

{{{(2b^2)=5a}}}

{{{a=(2b^2)/5}}}


go to =>{{{9=a^2+b^2}}}, substitute {{{a}}}

{{{9=((2b^2)/5)^2+b^2}}}.......solve for {{{b}}}

{{{9=(4b^4)/25+b^2}}}
{{{9*25=4b^4+25b^2}}}
{{{4b^4+25b^2-225=0}}}
{{{(b^2 - 5) (4 b^2 + 45) = 0}}}

solutions:

=>{{{b^2 =5}}}
or
 {{{b^2= -45/5}}}

Since, a is distance, it cant be negative.

using positive solution. find {{{a}}}

{{{a=(2*5)/5}}}

{{{a=2}}}=>{{{a^2=4}}}


and your equation is:


{{{x^2/4 - y^2/5=1 }}}



{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-3,0,.12),circle(3,0,.12),
locate(-3,0.5,F(-3,0)),locate(3,0.5,F(3,0)),
 graph( 600, 600, -10, 10, -10, 10, -(1/2)sqrt(5)*sqrt(x^2-4),(1/2)sqrt(5)*sqrt(x^2-4))) }}}