Question

which of the following expresses the coordinates of the foci of the conic section shown below?
(x-2)^2/4+(y+5)^2/9=1

Answer 1

@triciaal

Answer 2

@triciaal

Answer 3

ellipse right?

Answer 4

@triciaal yes an ellipse!

Answer 5

this should help
State the center, foci, vertices, and co-vertices of the ellipse with equation
25x2 + 4y2 + 100x – 40y + 100 = 0. Also state the lengths of the two axes.
I first have to rearrange this equation into conics form by completing the square and dividing through to get "=1". Once I've done that, I can read off the information I need from the equation.

25x2 + 4y2 + 100x – 40y = –100
25x2 + 100x + 4y2 – 40y = –100
25(x2 + 4x ) + 4(y2 – 10y ) = –100 + 25( ) + 4( )
25(x2 + 4x + 4) + 4(y2 – 10y + 25) = –100 + 25( 4 ) + 4( 25 )
25(x + 2)2 + 4(y – 5)2 = –100 + 100 + 100 = 100
25(x+2)^2/100 + 4(y-5)^2/100 = 100/100, so (x + 2)^2 / 4 + (y - 5)^2 / 25 = 1

The larger demoninator is a2, and the y part of the equation has the larger denominator, so this ellipse will be taller than wide (to parallel the y-axis). Also, a2 = 25 and b2 = 4, so the equation b2 + c2 = a2 gives me 4 + c2 = 25, and c2 must equal 21. The center is clearly at the point (h, k) = (–2, 5). The vertices are a = 5 units above and below the center, at (–2, 0) and (–2, 10). The co-vertices are b = 2 units to either side of the center, at (–4, 5) and (0, 5). The major axis has length 2a = 10, and the minor axis has length 2b = 4. The foci are messy: they're sqrt[21] units above and below the center.

center (–2, 5), vertices (–2, 0) and (–2, 10),
co-vertices (–4, 5) and (0, 5), foci (-2, 5 - sqrt[21]) and (-2, 5 + sqrt[21]),
major axis length 10, minor axis length 4

Answer 6

@triciaal thank you! all the info above is very helpful, but the answer you gave me is not one of my answer choices :(

Answer 7

Answer choices are
A. (2, -5+/-sqrt[13])
B. (2+/-sqrt[5],-5)
C. (2, -5+/-sqrt[5])
D. (2+/-sqrt[13],-5)

Answer 8

@triciaal

Answer 9

sorry don't know and cannot rush

Answer 10

If we make this traslation:
\[\Large \left\{ \begin{gathered}
x - 2 = X \hfill \\
y + 5 = Y \hfill \\
\end{gathered} \right.\]
where X and Y are the new coordinates, then we can rewrite your ellipse as below:
\[\Large \frac{{{X^2}}}{4} + \frac{{{Y^2}}}{9} = 1\]
Now the foci of that ellipse are the subsequent points:
\[\Large \begin{gathered}
{F_1} = \left( {0,\sqrt 5 } \right) \hfill \\
{F_2} = \left( {0, - \sqrt 5 } \right) \hfill \\
\end{gathered} \]
namely:
\[\Large {X_1} = 0,{Y_1} = \sqrt 5 \]
for F1, then we have to return to our old coordinate x,y, so using my trasformation above I get:
\[\Large \begin{gathered}
{x_1} = {X_1} + 2 = 0 + 2 = 0 \hfill \\
{y_1} = {Y_1} - 5 = - 5 + \sqrt 5 \hfill \\
\end{gathered} \]
so F1, written using teh old coordinates x,y, is:
\[\Large \left( {2, - 5 + \sqrt 5 } \right)\]
similarly we have for F2

Answer 11

the*

Answer 12

more explanation:
if we have the subsequent ellipse:
\[\Large \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\quad {b^2} > {a^2}\]
then the coordinates of it foci are:
\[\Large \begin{gathered}
{F_1} = \left( {0, - \sqrt {{b^2} - {a^2}} } \right) \hfill \\
{F_2} = \left( {0,\sqrt {{b^2} - {a^2}} } \right) \hfill \\
\end{gathered} \]

Answer 13

@Michele_Laino thank you so much! You were a great help :)

Answer 14

:)