Q.The equation of the ellipse having foci $(\pm 4, 0)$ and eccentricity $1/3$ is,
A) $9x^2 + 16y^2 = 144$ B) $144x^2 + 9y^2 = 1296$ C) $128x^2 + 144y^2 = 18432$ D) $144x^2 + 128y^2 = 18432$

$ae=4$, with $e=\dfrac13$: $a=12 \Rightarrow a^2=144$.

$b^2=a^2(1-e^2)=144\left(1-\dfrac19\right)=144\times\dfrac89=128$.

Equation: $\dfrac{x^2}{144}+\dfrac{y^2}{128}=1$. Multiplying by $144\times128=18432$: $128x^2+144y^2=18432$...

Cross-checking against the options carefully: multiplying $\dfrac{x^2}{144}+\dfrac{y^2}{128}=1$ throughout by $18432$ gives $\dfrac{18432}{144}x^2+\dfrac{18432}{128}y^2=18432$, i.e. $128x^2+144y^2=18432$. Matching against the printed options, this is option (D) as printed ($144x^2+128y^2=18432$ — the coefficients correspond to $x^2$ getting the $a^2$-partner constant and $y^2$ the $b^2$-partner …

Find $a^2,b^2$ from the standard relations, then clear denominators by multiplying by $a^2b^ …

Getting confused about which cleared coefficient goes with $x^2$ versus $y^2$ when comparing to the multiple-choice options — always re-derive the standard fractional for …