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anonymous
 one year ago
which of the following expresses the coordinates of the foci of the conic section shown below?
(x2)^2/4+(y+5)^2/9=1
anonymous
 one year ago
which of the following expresses the coordinates of the foci of the conic section shown below? (x2)^2/4+(y+5)^2/9=1

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@triciaal yes an ellipse!

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0this should help State the center, foci, vertices, and covertices 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(y5)^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 yaxis). 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 covertices 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), covertices (–4, 5) and (0, 5), foci (2, 5  sqrt[21]) and (2, 5 + sqrt[21]), major axis length 10, minor axis length 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@triciaal thank you! all the info above is very helpful, but the answer you gave me is not one of my answer choices :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Answer choices are A. (2, 5+/sqrt[13]) B. (2+/sqrt[5],5) C. (2, 5+/sqrt[5]) D. (2+/sqrt[13],5)

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0sorry don't know and cannot rush

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1If 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1more 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: \[\Large \begin{gathered} {F_1} = \left( {0,  \sqrt {{b^2}  {a^2}} } \right) \hfill \\ {F_2} = \left( {0,\sqrt {{b^2}  {a^2}} } \right) \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino thank you so much! You were a great help :)
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