Question 951534


{{{x^2-3y^2=144 }}}

{{{x^2/144-3y^2/144=144 /144}}}

{{{x^2/144-y^2/48=1}}}

center: ({{{0}}},{{{0}}})

{{{a^2=144}}} => {{{a=sqrt(144)=12}}}

{{{b^2=48}}} => {{{b=sqrt(48)=6.93}}}

then {{{c^2=a^2+b^2}}} =>{{{c^2=144+48=192}}} => {{{c=sqrt(192)}}}
=> {{{c=sqrt(64*3)}}}=> {{{c=8sqrt(3)}}}

The Latus Rectum is the line through the focus and parallel to the directrix.

The length of the Latus Rectum is: 

{{{2b^2/a=(2*48)/12=48/6=8}}}

positive end of latus rectum: ({{{L}}},{{{4}}})



positive end of latus rectum at Point ({{{L}}},{{{4}}}): 

({{{c }}} ,{{{b^2/a}}})=({{{8sqrt(3)}}} ,{{{4}}})

normal at ({{{8sqrt(3)}}} ,{{{4}}})=({{{x[1]}}},{{{y[1]}}}) is:

{{{a^2x/x[1]+b^2y/y[1]=a^2+b^2}}}

{{{144x/8sqrt(3)+48y/4=144+48}}}

{{{6sqrt(3)x+12y = 192}}} .......divide by 6

{{{sqrt(3)x+2y-32 =0}}}


{{{drawing( 600, 600, -50, 50, -50, 50,
circle(8sqrt(3),4,.52),locate(8sqrt(3),4,L(8sqrt(3),4)),
 graph( 600, 600, -50, 50, -50, 50,-sqrt((x^2/144-1)48),sqrt((x^2/144-1)48),-sqrt(3)x/2+16))) }}}