the foci of ((x-5)^2 /25) + ((y+3)^2 /9) = 1

Question

anonymous · Answer

(1, -3)  |  (9, -3)

anonymous · Answer

http://www.wolframalpha.com/input/?i=find+the+foci+of+%28%28x-5%29%5E2+%2F25%29+%2B+%28%28y%2B3%29%5E2+%2F9%29+%3D+1

anonymous · Answer

Does that answer your question?

apoorvk · Answer

Okay can you guess what changes "x-5" and "y+3" cause in 
$$\frac{(x-5)^2}{25} + \frac{(y+3)^2}{9} = 1$$
I mean what is their significance? Because the equation of a standard ellipse is :
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

Right? So what changes are those incorporating into this?

Try to think on the lines of a circle's equation if you've studied them...

anonymous · Answer

thanks

apoorvk · Answer

Hmm, already understood that?

anonymous · Answer

yes..i think i do ..

apoorvk · Answer

Brilliant then! still for my own satisfaction, can you answer my question above?