Question 932086
{{{49x^2 + y^2 = 49}}}........both sides divide by {{{49}}}


{{{49x^2/49 + y^2/49 = 49/49}}}

{{{x^2/1 + y^2/49 = 1}}} => you have an ellipse; compare it to the standard form equation of an ellipse which is {{{(x-h)^2/a^2 + (y-k)^2/b^2 = 1}}} (your  example is an ellipse with Vertical Major Axis,because {{{b>a}}}) where ({{{h}}},{{{k}}}) is the center, {{{a}}} is semiminor axis length, {{{b}}} is semimajor axis length

Ellipses are actually very special cases of circles. 

REMEMBER: 
The right side of the equation must be {{{1}}} in order to be in standard form.
The point ({{{h}}},{{{k}}}) is most commonly referred to as the CENTER.
In order to graph an ellipse, you need {{{4}}} points:
Right point ({{{h+a}}},{{{k}}})
Left point ({{{h-a}}},{{{k}}})
Top point ({{{h}}},{{{k+b}}})
Bottom point ({{{h}}},{{{k-b}}}). 


in your case:

center at ({{{0}}},{{{0}}})
foci | (({{{0}}},{{{ -4sqrt(3)}}})  |  ({{{0}}}, {{{4sqrt(3)}}}))approx.(({{{0}}}, {{{-6.9}}})  |  ({{{0}}},{{{ 6.9}}}))
vertices | ({{{0}}},{{{ -7}}})  |  ({{{0}}], {{{7}}})
semimajor axis length | {{{7}}}
semiminor axis length |{{{ 1}}}
eccentricity | ({{{4sqrt(3))/7}}}) approx.{{{0.989743}}}



{{{ graph( 600, 600, -10,10, -10, 10, sqrt(49-49x^2),-sqrt(49-49x^2)) }}}