Question

given the ellipse with foci at (5,3) and (5,-5) and a major axis of length 12, determine the center and the denominator of the equation
equation:
(x ?)^2/? + (y ?)^2/?=1

Answer 1

so... where do you think the center might be? keep in mind that the center is half-way for either major or minor axises

Answer 2

5,-2?

Answer 3

well... I was going to say... is also half-way of the foci distances

so half-way between 3 and -5 over the y-axis.. is?

Answer 4

-1?

Answer 5

well... . don't forget the x-coordinate... so yes.. is (5, -1) so the center is there

so the focus are one up and another below the center
that means is a vertical ellipse
that means the major axis, or the "a" component will be UNDER the fraction with the "y" in it

also keep in mind that the major axis is given, is 12 and that

Answer 6

so to find the minor axis from here i would set it up like this: 36-b^2=25?

Answer 7

\(\bf \cfrac{(x-h)^2}{b^2}+\cfrac{(y-k)^2}{a^2}=1\implies \cfrac{(x-{\color{brown}{ 5}})^2}{b^2}+\cfrac{(y-{\color{brown}{ (-1)}})^2}{{\color{brown}{ 6}}^2}=1\)

so.. what's "b"? well.. we dunno
but we know that the distance from the center (5,-1) to either focus is 4 units
that is a distance of "c"
keep in mind that \(\bf c=\sqrt{{\color{brown}{ a}}^2-{\color{blue}{ b}}^2}\implies \sqrt{c^2+{\color{brown}{ a}}^2}={\color{blue}{ b}}\)

Answer 8

so your saying c=4 so b would be sqrt(52) when i put it into the equation b=sqrt(c^2+a^2)?

Answer 9

hmmm hold.. on.. .lemme recheck ... that.. I get a number too big for "b"

Answer 10

hmmm I see.. the issue

Answer 11

so the "c" distance should be -> \(\bf c=\sqrt{{\color{brown}{ a}}^2-{\color{blue}{ b}}^2}\implies c^2={\color{brown}{ a}}^2-{\color{blue}{ b}}^2\implies {\color{blue}{ b}}^2={\color{brown}{ a}}^2-c^2\implies {\color{blue}{ b}}=\sqrt{{\color{brown}{ a}}^2-c^2}\)

Answer 12

so b would equal 2sqrt(5)

Answer 13

yeap

Answer 14

thus \(\bf \cfrac{(x-5)^2}{{\color{blue}{(2\sqrt{5}) }}^2}+\cfrac{(y+1)^2}{{\color{brown}{ 6}}^2}=1\implies \cfrac{(x-5)^2}{{\color{blue}{20 }}}+\cfrac{(y+1)^2}{{\color{brown}{ 36}}}=1\)

Answer 15

ok thank you very much

Answer 16

yw