Vertices/Foci of an ellipse

**Greymalkin** · August 20th 2012, 12:34 PM

Should be an easy one: Find foci and vertices
4x²+9y²=1

answers are
Vertices= $({{\frac{-1}{2}},0})({{\frac{1}{2}},0})$

Foci= $({\frac{-\sqrt{5}}{6}},0)({\frac{\sqrt{5}}{6}},0)$

I cant seem to be able to turn the equation into proper ellipse form to derive numbers from.

**Soroban** · August 20th 2012, 01:31 PM

Hello, Greymalkin!

$\text{Find foci and vertices: }\:4x^2 +9y^2 \:=\:1$

$\text{Answers are:}$

. . $\text{Vertices: }\left(\pm\tfrac{1}{2},\,0\right)$

. . . $\text{Foci: }\left(\pm\tfrac{\sqrt{5}}{6},\,0\right)$

$\text{I can't seem to be able to turn the equation into proper ellipse form to derive numbers.}$

The general form is: . $\frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1$

Note the coefficient of $x^2$ should be 1.

We have: . $4x^2 + 9y^2 \:=\:1$

Multiply the first term by $\frac{\frac{1}{4}}{\frac{1}{4}}$ , the second by $\frac{\frac{1}{9}}{\frac{1}{9}}$

. . $\frac{\frac{1}{4}}{\frac{1}{4}}\left(\frac{4x^2}{1 }\right) + \frac{\frac{1}{9}}{\frac{1}{9}}\left(\frac{9y^2}{1 }\right) \:=\:1 \quad\Rightarrow\quad \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{9}} \:=\:1$

Hence: . $\begin{Bmatrix}a^2 \:=\:\frac{1}{4} & \Rightarrow & a \:=\:\pm\frac{1}{2} \\ \\[-3mm] b^2 \:=\:\frac{1}{9} & \Rightarrow & b \:=\:\pm\frac{1}{3} \end{Bmatrix}$

And: . $c^2 \:=\:a^2-b^2 \:=\:\tfrac{1}{4} - \tfrac{1}{9} \:=\:\tfrac{5}{36} \quad\Rightarrow\quad c \:=\:\pm\tfrac{\sqrt{5}}{6}$

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