Question 170009
{{{49x^2 + 16y^2 = 784}}} Start with the given equation



{{{(49x^2 + 16y^2)/784 = cross(784/784)}}} Divide both sides by 784. The goal here is to make the right side equal to 1



{{{(49x^2 + 16y^2)/784 = 1}}} Simplify



{{{(49x^2)/784 + (16y^2)/784 = 1}}} Break up the fraction



{{{(x^2)/16 + (y^2)/49 = 1}}} Reduce



{{{(x^2)/(4^2) + (y^2)/(7^2) = 1}}} Rewrite 16 as {{{4^2}}}. Rewrite 49 as {{{7^2}}}



Now the equation is in the form {{{((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) = 1}}} where {{{h=0}}}, {{{k=0}}}, {{{a=4}}}, and {{{b=7}}}.



The values of "a" and "b" are the lengths of the major and minor axis. If "a" is larger than "b", then {{{2a}}} is the length of the major axis. On the other hand, if "b" is larger than "a", then {{{2b}}} is the length of the major axis.



Since {{{b>a}}}, this means that the length of the major axis is {{{2b}}}.



So multiply the value of "b" by 2 to get {{{2*7=14}}}




So the length of the major axis is 14 units