Focal Radii

Distances from P(x₁, y₁) on the ellipse to the two foci S(−c, 0) and S′(c, 0). r₁ is the distance to the nearer focus when x₁ > 0. Both are positive since −a ≤ x₁ ≤ a.

For the ellipse $x^2/a^2 + y^2/b^2 = 1$ with foci $S(-c, 0)$ and $S'(c, 0)$, let $P(x_1, y_1)$ be on the ellipse.

The corresponding directrices are $x = -a/e$ (for $S$) and $x = a/e$ (for $S'$).

Using the focus-directrix property $PS/PM = e$:

Distance to $S'(c, 0)$: The directrix corresponding to $S'$ is $x = a/e$.

$$PM' = \frac{a}{e} - x_1 \quad (x_1 \leq a, \text{ so } PM' > 0)$$ $$r_2 = PS' = e \cdot PM' = e\!\left(\frac{a}{e} - x_1\right) = a - ex_1$$

Distance to $S(-c, 0)$: The directrix corresponding to $S$ is $x = -a/e$.

$$PM = x_1 - \left(-\frac{a}{e}\right) = x_1 + \frac{a}{e}$$ $$r_1 = PS = e \cdot PM = e\!\left(x_1 + \frac{a}{e}\right) = ex_1 + a$$

Conventionally (taking $S'$ as the nearer focus for $x_1 > 0$):

$$r_1 = a - ex_1 \quad \text{(to nearer focus)}, \quad r_2 = a + ex_1 \quad \text{(to farther focus)}$$

Since $-a \leq x_1 \leq a$ and $0 < e < 1$: both $r_1 = a - ex_1 \geq a - ea = a(1-e) > 0$ and $r_2 = a + ex_1 \geq a - ea > 0$. Both focal radii are always positive.