Question

Find the center, vertices, co-vertices,focal points of
y^2/36+x^2/4 = 1.
I got 0, (6,-6) and 4,(the sq root of 2)

Answer 1

I got 0, (6,-6) <--- resembles closeness
did you mean center at (0, 0), and vertices at .. (\(\pm\) 6, 0) ?

Answer 2

\[\frac{y^2}{36}+\frac{x^2}{4} = 1\]

here is the form of an ellipse:
\[\frac {(y-k)^2}{a^2} + \frac {x-h)^2}{b^2} = 1\] the point (h,k) is the vertex, '
now looking at your own equation, what would you say h and k equal?


[\frac {(y-k)^2}{a^2} + \frac {x-h)^2}{b^2} = 1\]

Answer 3

\[\frac {(y-k)^2}{a^2} + \frac {(x-h)^2}{b^2} = 1\]

[\frac {(y-k)^2}{a^2} + \frac {(x-h)^2}{b^2} = 1\]

Answer 4

sorry I was a bit ... caught up... so
\(\bf \cfrac{y^2}{36}+\cfrac{x^2}{4}=1\implies \cfrac{(y-0)^2}{6^2}+\cfrac{(x-0)^2}{2^2}=1\\ \quad \\\implies \cfrac{(y-k)^2}{a^2}+\cfrac{(x-h)^2}{b^2}=1\\ \quad \\
\textit{center is at (h, k)}\\ \quad \\
\textit{vertices are at }(h, k\pm a)\\ \quad \\
\textit{co-vertices are at }(h\pm b, k)\\ \quad \\
\textit{focal points are at }c=\sqrt{a^2-b^2}\implies (h,k\pm c)\)