Question 117842
Given:
 The equation of an ellipse (with horizontal major axis) in standard form: {{{(x-h)^2/a^2 + (y-k)^2/b^2=1}}} 

 {{{x^2/525^2 + y^2/350^2=1}}}


From given equation, we can say that this is an ellipse with {{{a = 526}}} and {{{b = 350}}} (note: {{{a >b}}}), and has the center in the origin (0,0) since the {{{h=0}}} and {{{k=0}}}
 	

{{{1 - x^2/525^2 = y^2/350^2}}}…………..multiply both sides by {{{350^2}}}

{{{(-x^2/525^2) 350^2+ (1*350^2) = y}}}
	

{{{-(x ^2)(350^2/525^2) + 350^2=y}}}……………………………………………


Since {{{350^2/525^2=350*350/(525*525)=(350:175)(350:175)/((525:175)(525:175))= 2*2/(3*3)=4/9}}}, we have


{{{-x^2 (4/9) + 350^2=y}}}……………
	

Set {{{y = 0}}} in the equation obtained and find the{{{ x –intercepts}}}.	


{{{-x^2 (4/9) + 350^2= 0 }}}……………move {{{-x^2 (4/9)}}} to the right


{{{350^2 = x^2 (4/9)}}}………divide both sides by {{{4/9}}}


{{{350^2/(4/9) = x^2 *(4/9)/( 4/9)}}}………


{{{9*350^2/4= x^2 }}}………

	
{{{9*122500/4= x^2 }}}………


{{{9*30625= x^2 }}}………


{{{275625= x^2 }}}……… 


{{{x[1,2] = sqrt (275625))}}}………..

So,
{{{x[1] = 525}}}

{{{x[2] = -525}}}