Question

Find the center, vertices, and foci of the ellipse with equation x^2/100+y^2/36= 1

Answer 1

\[ \large
\frac{x^2}{100} + \frac{y^2}{36} = 1 \\ \large
\frac{x^2}{10^2} + \frac{y^2}{6^2} = 1 ~~~~ ----- (1)\\ \large
\]Compare this to th general form\[
\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 ~~~~ ----- (2) \\ \large
\]where (h,k) is the center, a is the semi=major axis and b is the semi-minor axis.

Can you find (h,k), a and b bu comparing (1) and (2)?

Answer 2

a= 10 b=6 but how do you get h and k when you only have x and y without anything in front of it

Answer 3

@aum

Answer 4

x^2/100+y^2/36= 1,
So x^2/10^2+y^2/6^2= 1,
By this standard form of ellipse function, the center is (0,0);
And a=10>b=6,
Since a coordinates with x, so the foci are on the X axis,
Since a^2=b^2+c^2, c=8,
So foci are (-8,0) and (8,0)
By definition of vertices, let y=0, so x=10 or -10 =|a|, so vertices are (10,0) and (-10,0)

Answer 5

thank you so much @WilliamZhang !