Conic Sections

[hgaon001]

Im told to find the center, vertices, foci, and eccentricity of the conic section given by this equation: 25x^2 + 9y^2 + 100x -72y + 19 = 0

i split the y's and x's then i complete the square for each.. so at the end i get

[9(y-4)^2]+[25(x+2)^2]=1

i believe this is an ellipse.. but im lost on how to graph it because i thought the standard formula of an ellipse would be in form of [(x^2)/(a^2)] + [(y^2)/(b^2)] = 1

 

[Deleted member 4993]

Im told to find the center, vertices, foci, and eccentricity of the conic section given by this equation: 25x^2 + 9y^2 + 100x -72y + 19 = 0

i split the y's and x's then i complete the square for each.. so at the end i get

[9(y-4)^2]+[25(x+2)^2]=1

i believe this is an ellipse.. but im lost on how to graph it because i thought the standard formula of an ellipse would be in form of [(x^2)/(a^2)] + [(y^2)/(b^2)] = 1

[9(y-4)^2]+[25(x+2)^2]=1

9y[sup:2pf9rtl9]2[/sup:2pf9rtl9] - 72y + 144 + 25x[sup:2pf9rtl9]2[/sup:2pf9rtl9] + 100x + 100 = 1

25x[sup:2pf9rtl9]2[/sup:2pf9rtl9] + 9y[sup:2pf9rtl9]2[/sup:2pf9rtl9] + 100x - 72y + 243 = 0

Is it equivalent to

25x^2 + 9y^2 + 100x -72y + 19 = 0

I wouldn't say so.

 

[hgaon001]

thanks! i see what i did wrong

 

[BigGlenntheHeavy]

\(\displaystyle You \ should \ get \ \frac{(x+2)^{2}}{3^{2}}+\frac{(y-4)^{2}}{5^{2}} = 1\)
[attachment=0:hb5pjedl]wer.jpg[/attachment:hb5pjedl]

Here is its graph.