Question

I WILL GIVE MEDAL AND FAN:
Find the center, vertices, and foci of the ellipse with equation ((x^2/)/(100)) + ((Y^2)/(64))= 1

Answer 1

@nincompoop

Answer 2

@ash2326

Answer 3

@ganeshie8

Answer 4

standard ellipse equation
For a wider-than-tall ellipse with center at (h, k), having vertices a units to either side of the center and foci c units to either side of the center, the ellipse equation is:
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]
For a taller-than-wide ellipse with center at (h, k), having vertices a units above and below the center and foci c units above and below the center, the ellipse equation is:
\[\frac{(y-h)^2}{a^2}+\frac{(x-k)^2}{b^2}=1\]
standard foci equation \[a^2-c^2=b^2\]
standard center \[(h,k)\]

Answer 5

based on your equation
\[\frac{x^2}{100}+\frac{y^2}{64}=1\]
a > b
there's a horizontal major axis

Answer 6

EXAMPLE:

suppose we have the equation

\[\frac{x^2}{4}+\frac{y^2}{1}=1\]

Answer 7

since a > b we have a horizontal major axis
so we re-write this equation as

where a = 2 and b = 1
\[\frac{x^2}{2^2}+\frac{y^2}{1^2}=1\]

Answer 8

before we start graphing, we want the center. Recall that the standard ellipse equation is
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \]

For this equation we just have \[x^2 \] and \[y^2 \]

\[\frac{x^2-0}{2^2}+\frac{y^2-0}{1^2}=1\]

which means that our center \[(h,k) \rightarrow (0,0)\]

Answer 9

for vertices we need (-a,0), (a,0), (-b,0),(0,b)
since a = 2 and b = 1
our vertex points are
(-2,0) (2,0) (-1,0)(0,1)