Question 1043924
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find the point of intersection of x^2+y^2=25,
x^2/18+ y^2/32=1
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{{{x^2 + y^2}}} = 25,   (1)
{{{x^2/18 + y^2/32}}} = 1.    (2)

Multiply (2) by 18 (both sides). You will get

{{{x^2 + y^2}}} = 25,        (1')
{{{x^2 + y^2(18/32)}}} = 18,   (2')

or equivalently

{{{x^2 + y^2}}} = 25,        (1'')
{{{x^2 + y^2(9/16)}}} = 18.   (2'')

Distract (2'') from (1''). You will get

{{{y^2(1-9/16)}}} = 25 - 18,  or  {{{(7/16)y^2}}} = 7,  or {{{y^2}}} = 16.

Hence, y = +/- 4.

Then from (1) x = +/-3.

<U>Answer</U>. There are four solution and, respectively, four intersection points:
        (x,y) = (3,4), (3,-4), (-3,-4), (-3, 4).
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