Question 30338
i believe that the answer is b) except you have written the answer incorrectly.


{{{ x^2 + y^2 = 16 }}} and {{{ (x^2/25) + (y^2/16) = 1}}}


{{{ x^2 + y^2 = 16 }}}
{{{ y^2 = 16 - x^2 }}}. Sub this into the second equation, to give:


{{{ (x^2/25) + ((16 - x^2)/16) = 1 }}}
{{{ (x^2/25) + (16/16) - (x^2/16) = 1 }}}
{{{ (x^2/25) + 1 - (x^2/16) = 1 }}}
{{{ (x^2/25) - (x^2/16) = 0 }}}
{{{ -9x^2/400 = 0 }}}
--> {{{ x^2 = 0 }}}
so x = 0.


If x=0, then from {{{ y^2 = 16 - x^2 }}}, we get {{{ y^2 = 16 - 0^2 }}}
--> {{{ y^2 = 16 }}}
--> {{{ y = sqrt(16) }}} or {{{ y = -sqrt(16) }}}


so, y = 4 or -4.


The 2 solutions are (0,4) and (0,-4).


Check in the second equation, {{{ (x^2/25) + (y^2/16) = 1}}}


using (0,4):
{{{ (0^2/25) + (4^2/16) }}}
{{{ 0 + (16/16) }}}
which is indeed 1... correct.


using (0,-4):
{{{ (0^2/25) + ((-4)^2/16) = 1}}}
{{{ 0 + (16/16) = 1}}}
which is also 1... correct


jon.