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I have the following equation of an ellipse: $$5x^2+9y^2+40x=100$$

I need to put it in the form: $$\frac{(x-h)^2}{m^2} + \frac{(y-k)^2}{n^2} =1 $$

I was trying to complete the square with the coefficients that have an $x$ variable.

$$5x^2 + 40x + 9y^2 = 100$$ $$5(x^2 + 8x) + 9y^2 = 100$$ $$5(x^2 + 8x + 16) + 9y^2 = 100 + 16$$ $$5(x+4)^2 + 9y^2 = 116$$

I then divided both sides of the equation by $116$.

$$\frac{5(x + 4)^2}{116} + \frac{9y^2}{116} = 1$$

However, when I graph the equation I do not get the same ellipse that was represented with the original equation of $5x^2 + 9y^2 + 40x = 100$. From the graph, it seems like they are two similar ellipses. Where am I making the mistake? Any help will be appreciated.

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    $\begingroup$ You haven't added $16$ on the left hand side, you've added $5 \times 16 = 80$. $\endgroup$
    – user296602
    Apr 20, 2018 at 21:48
  • $\begingroup$ I see! Thank you for pointing our the error! $\endgroup$
    – geo_freak
    Apr 20, 2018 at 21:49

2 Answers 2

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You did a good job just you made a mistake here $$5(x^2 + 8x + 16) + 9y^2 = 100 + 16$$ Should be $$5(x^2 + 8x + 16) + 9y^2 = 100 + 80$$ $$5(x^2 +4)^2 + 9y^2 = 180$$ $$\frac {(x^2 +4)^2}{36} + \frac {y^2}{20} = 1$$ $$.....$$

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We have

$$5x^2+9y^2+40x=100\iff 5(x+4)^2+9y^2=180\iff (x+4)^2+\frac{y^2}{\frac59}=36\\\iff\frac{(x+4)^2}{36}+\frac{y^2}{20}=1$$

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  • $\begingroup$ When you divided by 5, you divided the $5(x+4)^2$ term by 5 twice. $\endgroup$ Apr 20, 2018 at 23:43
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    $\begingroup$ @OscarLanzi oh yes of course! Thanks $\endgroup$
    – user
    Apr 21, 2018 at 7:09

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