Ellipse equation equal to zero, how to define foci?

Question

I will appreciate if you could help me with Equation $10x^2+y^2+200x=0$ . The answer key is giving foci $(-10, 30)$ and $(-10,-30).$ I am confused: ellipse equation must be equal to $1.$ Or it is in a form of a line? I tried to make it standard form: $(10(x+10)^2)/(1/\sqrt {10})^2)+y^2=0$. But this zero...And where did $30$ and $-30$ come from? P.S. Sorry for the formating, this is my first time using this web-site, I couldnt figure how to put square root sign.

Answer 1

Completing squares we get $$10x^2+y^2+200x=0\iff (\sqrt{10}x+10\sqrt{10})^2+y^2=1000.$$

We write $1$ on the rigth hand side $$(\sqrt{10}x+10\sqrt{10})^2+y^2=1000\iff \dfrac{(\sqrt{10}x+10\sqrt{10})^2}{1000}+\dfrac{y^2}{1000}=1.$$

Simplifying we obtain $$\dfrac{(x+10)^2}{10^2}+\dfrac{y^2}{(10\sqrt{10})^2}=1.$$

Answer 2

HINT:

Divide by 10 and add 100 on both sides

$$\dfrac{(x+10)^2}{1^2}+\left( \dfrac{y}{\sqrt {10}} \right) ^2 =100$$ Divide by $100 =10^2$ $$\dfrac{(x+10)^2}{10^2}+\left( \dfrac{y}{10\sqrt {10}} \right) ^2 =1.$$

$$ a=10; b=10\sqrt{10};c=\sqrt{a^2-b^2}=30;$$

Push the ellipse to the left by 10 units along $x-$axis.