Conic Sections: Numerical Problems Set (11)

Question

Find the equation of the ellipse with foci at $(\pm 4, 0)$ and a directrix at $x = \frac{25}{4}$.

Solution — Step by Step

The equation: $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$.

Why This Works

An ellipse is fully determined by two focal points and any one directrix (or by other compatible pairs of data). The focal distance gives $ae$, the directrix gives $a/e$, and multiplying them eliminates $e$ and yields $a^2$.

Once $a$ and $e$ are known, $b^2 = a^2(1-e^2)$ closes the equation. The relationship $b^2 = a^2 - c^2$ also works as a cross-check.

Alternative Method

Use $b^2 = a^2 - c^2$ directly. We need $a$. From $a/e = 25/4$ and $c = 4$, $a/e = a^2/c \implies a^2 = 25c/4 \times \ldots$ — actually slower. Stick with the multiplication trick.

Common Mistake

Confusing the formula with a hyperbola, where $b^2 = c^2 - a^2$ instead. The fix: check whether $a > c$ (ellipse) or $a < c$ (hyperbola) before substituting. JEE Main 2024 had a four-mark trap on exactly this interchange.