alleryAn ellipse has foci at $(\pm 4, 0)$ and passes through the point $(5, 0)$. Find its standard equation.
A$\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$
B$\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$
C$\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$
D$\dfrac{x^2}{9} + \dfrac{y^2}{16} = 1$
Answer & Solution
Correct answer: A. $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$
**Read the geometry.** Foci on the $x$-axis at $(\pm 4, 0)$
⇒ horizontal ellipse with $c = 4$. The point $(5, 0)$ lies on the major axis ($y = 0$) and on the curve
⇒ it must be a *vertex*, giving $a = 5$.
**Recover $b$.** $b^2 = a^2 - c^2 = 25 - 16 = 9$, so $b = 3$.
**Equation:** $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$.
**Distractor check.** Option B has $b = 4$ — that would put the foci at $(\pm 3, 0)$, not the given $(\pm 4, 0)$.
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