The ellipse $\displaystyle 14x^2+2x+y^2 = 1 $ has its center at the point (b, c) where
$\displaystyle b = $
$\displaystyle c = $
The length of the major diameter of this ellipse is __
$\displaystyle 14x^{2}+2x+y^{2}=1$
Complete the square and get:
$\displaystyle 14(x+\frac{1}{14})^{2}+y^{2}=\frac{15}{14}$
$\displaystyle \frac{196(x+\frac{1}{14})^{2}}{15}+\frac{14y^{2}}{ 15}=1$
To find the x intercepts, set y=0 and solve for x, getting $\displaystyle x=(\frac{-(\sqrt{15}+1)}{14},\frac{\sqrt{15}-1}{14})$
The center is at (-1/14,0)
$\displaystyle b=\frac{-1}{14}+\frac{\sqrt{15}-1}{14}=\frac{\sqrt{15}}{14}$
Setting y=-1/14 in the equation gives us major axes of $\displaystyle a=(\frac{\sqrt{210}}{14}, \frac{-\sqrt{210}}{14})$
The foci, c, can be found by $\displaystyle c^{2}=a^{2}-b^{2}$
$\displaystyle c=\sqrt{(\frac{\sqrt{210}}{14})^{2}-(\frac{\sqrt{15}}{14})^{2}}=\pm\frac{\sqrt{195}}{1 4}$