# Given the ellipse defined by the equation, find the length of the major axis and the minor axis and find the eccentricity. 36x^2+9y^2+72x-36y-2552=0

Tony Burgess 2022-09-14 Answered
Given the ellipse defined by the equation, find the length of the major axis and the minor axis and find the eccentricity.
$36{x}^{2}+9{y}^{2}72x-36y-252=0$
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## Answers (1)

Aubrie Conley
Answered 2022-09-15 Author has 13 answers
Answer:
$36{x}^{2}+9{y}^{2}72x-36y-252=0\phantom{\rule{0ex}{0ex}}36\left({x}^{2}+2x+1\right)-36+9\left({y}^{2}-4y+4\right)-252=0\phantom{\rule{0ex}{0ex}}36\left(x+1{\right)}^{2}+9\left(y-2{\right)}^{2}=324\phantom{\rule{0ex}{0ex}}\frac{\left(x+1{\right)}^{2}}{\left(324/36\right)}+\frac{\left(y-2{\right)}^{2}}{\left(324/9\right)\right)}=1\phantom{\rule{0ex}{0ex}}\frac{\left(x+1{\right)}^{2}}{{3}^{2}}+\frac{\left(y-2{\right)}^{2}}{{6}^{2}}=1$
Here, comparing with standard eclipse
eccentricity $=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{1-\left(\frac{3}{6}{\right)}^{2}}\phantom{\rule{0ex}{0ex}}\sqrt{1-1/4}=\frac{\sqrt{3}}{2}$

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