Question 1133039

Standard form:
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}->Transverse axis is horizontal

{{{(y-h)^2/a^2-(x-k)^2/b^2=1}}}-.>Transverse axis is vertical


1) 

given:

Vertices: ({{{0}}},{{{-6}}}), ({{{0}}},{{{6}}}) =>Semi-major axis length: {{{a=6}}}

 {{{(x-h)^2/6^2-(y-k)^2/b^2=1}}}

 {{{(x-h)^2/36-(y-k)^2/b^2=1}}}

First asymptote: {{{y=-5x}}}
Second asymptote: {{{y=5x}}}

=> Transverse axis is vertical

Standard form: {{{(y-h)^2/a^2-(x-k)^2/b^2=1}}}

intersection of the asymptotes is at origin: 

so, center is at: ({{{0}}},{{{0}}}) => {{{h=0 }}}and {{{k=0}}}

=>{{{y^2/a^2-x^2/b^2=1}}}

From the original equations of asymptotes, you can determine the slopes of the asymptotes to be {{{m=5}}} and {{{m=-5}}}

since {{{m=a/b}}} and {{{a=6}}},  we have

{{{5=6/b}}}

=>{{{b=6/5}}}

{{{x^2/36-y^2/(6/5)^2=1}}}

{{{x^2/36-y^2/(36/25)=1}}}

{{{x^2/36-25y^2/36=1}}}







2.  



2) focos (0,+-6) pasa por P=(-5,9)


the coordinates of the foci are ({{{0}}},±{{{c}}})

foci lie on y axis, so your formula is:

{{{y^2/a^2-x^2/b^2=1}}}

=> {{{c=6}}} or {{{c=-6}}}

{{{6^2=a^2+b^2}}}
{{{36=a^2+b^2}}}
{{{b^2=36-a^2}}}

 P=({{{-5}}},{{{9}}})

 *[invoke formula_distance 0, -6, -5, 9]

{{{d[1]=15.8113883008419}}}=> round it to {{{d[1]=16}}}

*[invoke formula_distance 0, 6, -5, 9]

{{{d[2]=5.8309518948453}}}=> round it to {{{d[2]=6}}}

{{{d[1]-d[2]=2a}}}

{{{2a=16-6=10}}}

{{{a=10/2=5}}}

{{{b^2=36-5^2}}}
{{{b^2=36-25}}}
{{{b^2 = 11}}}

{{{y^2/25-x^2/11=1}}}


3) focos (0,+-1) longitud eje real:1

longitud eje real:1=> the length transverse axis is {{{1}}}

=>{{{2a=1}}}
=>{{{a=1/2}}}

the coordinates of the foci are ({{{0}}},±{{{c}}}) => {{{c}}}=±{{{1}}} and

foci lie on {{{y}}} axis, so your formula is

{{{y^2/a^2-x^2/b^2=1}}}

=> {{{c=1}}} or {{{c=-1}}}

{{{1^2=(1/2)^2+b^2}}}

{{{1=1/4+b^2}}}

{{{b^2=1-1/4}}}

{{{b^2=3/4}}}

{{{y^2/(1/4)-x^2/(3/4)=1}}}

{{{4y^2-4x^2/3=1}}}


4) asíntotas {{{y}}}= ± {{{x/2}}} pasa por el punto de coordenadas (5,2) 



{{{y}}}= ± {{{(1/2)x}}}

intersection of the asymptotes is at origin: 

so, center is at: ({{{0}}},{{{0}}}) => {{{h=0 }}}and {{{k=0}}}

 asymptotesup-down {{{y}}}= ± {{{(a/b)x}}}

{{{a/b=1/2}}}
{{{2a=b}}}

since passes through ({{{5}}},{{{2}}}) 

{{{5^2/a^2-2^2/b^2=1}}}

{{{25/a^2-4/(2a)^2=1}}}

{{{25/a^2-4/4a^2=1}}}

{{{25/a^2-1/a^2=1}}}

{{{24/a^2=1}}}

{{{a^2=24}}}

{{{b=2a=2sqrt(24)}}}

{{{b^2=4*24}}}

{{{b^2=96}}}


{{{y^2/24-x^2/96=1}}}