Q.Find the

$3x^2+4y^2=1 \Rightarrow \dfrac{x^2}{1/3}+\dfrac{y^2}{1/4}=1$. So $a^2=\dfrac13,\,b^2=\dfrac14 \Rightarrow a=\dfrac{\sqrt3}{3},\,b=\dfrac12$.

1. Major axis $=2a=\dfrac{2\sqrt3}{3}$; minor axis $=2b=1$.
2. $e^2=1-\dfrac{b^2}{a^2}=1-\dfrac{1/4}{1/3}=1-\dfrac34=\dfrac14 \Rightarrow e=\dfrac12$. Foci $(\pm ae,0)=\left(\pm\dfrac{\sqrt3}{6},0\right)$.
3. Directrices: $x=\pm\dfrac{a}{e}=\pm\dfrac{2\sqrt3}{3}$.
4. Latus rectum $=\dfrac{2b^2}{a}=\dfrac{2(1/4)}{\sqrt3/3}=\dfrac{1/2}{\sqrt3/3}=\dfrac{\sqrt3}{2}$.
5. Distance between foci $=2ae=\dfrac{\sqrt3}{3}$. …

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Losing track of the fractions $\frac13,\frac14$ and treating them as if the equ …