Question 612542
The standard form for the equation of an ellipse is
{{{(x-h)^2/a^2 + (y-k)^2/b^2 = 1}}}
Since the intercepts are on the ellipse in question, they must fit the equation. Substituting (5, 0) into this equation we get:
{{{(5-h)^2/a^2 + (0-k)^2/b^2 = 1}}}
which simplifies to:
{{{(25-10h+h^2)/a^2 + k^2/b^2 = 1}}}
Subtracting {{{k^2/b^2}}} from each side:
{{{(25-10h+h^2)/a^2 = 1 - k^2/b^2}}}
Multiplying by {{{a^2}}}:
{{{25-10h+h^2 = a^2(1 - k^2/b^2)}}}<br>
Substituting (-5, 0) into {{{(x-h)^2/a^2 + (y-k)^2/b^2 = 1}}} and following the same steps as above we end up with:
{{{25+10h+h^2 = a^2(1 - k^2/b^2)}}}<br>
Since {{{25-10h+h^2}}} and {{{25+10h+h^2}}} are both equal to {{{a^2(1 - k^2/b^2)}}}, they must be equal to each other:
{{{25-10h+h^2 = 25+10h +h^2}}}
We can solve this for h. Subtracting 25 and {{{h^2}}} and adding 10h we get:
0 = 20h
Dividing by 20:
0 = h<br>
Using (0, 3) and (0, -3) and logic just like the above we can find that k = 0.<br>
This makes the center of the ellipse (0, 0).<br>
Since the center, (5, 0) and (-5, 0) are all on the x-axis, the given points are not just the intercepts, they are the vertices on the major axis of the ellipse. This means the "a" is 5.<br>
Since the center, (0, 3) and (-5, 0) are all on the y-axis, the given points are not just the intercepts, they are the vertices on the minor axis of the ellipse. This means the "b" is 3.<br>
With the center, a and b we can now write the equation:
{{{(x-0)^2/5^2 + (y-0)^2/3^2 = 1}}}
which simplifies to:
{{{x^2/25 + y^2/9 = 1}}}