## anonymous one year ago 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

1. anonymous

@triciaal

2. anonymous

@triciaal

3. triciaal

ellipse right?

4. anonymous

@triciaal yes an ellipse!

5. triciaal

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

6. anonymous

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

7. anonymous

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

8. anonymous

@triciaal

9. triciaal

sorry don't know and cannot rush

10. Michele_Laino

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

11. Michele_Laino

the*

12. Michele_Laino

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: $\Large \begin{gathered} {F_1} = \left( {0, - \sqrt {{b^2} - {a^2}} } \right) \hfill \\ {F_2} = \left( {0,\sqrt {{b^2} - {a^2}} } \right) \hfill \\ \end{gathered}$

13. anonymous

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

14. Michele_Laino

:)