Eccentricity of ellipse x²/25 + y²/16 = 1 and its foci

Question

Find the eccentricity and foci of the ellipse:

\frac{x^2}{25} + \frac{y^2}{16} = 1

(NCERT Class 11, Exercise 11.3)

Solution — Step by Step

Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ :

$a^2 = 25 \Rightarrow a = 5$ and $b^2 = 16 \Rightarrow b = 4$ .

Since $a > b$ , the major axis is along the x-axis.

c^2 = a^2 - b^2 = 25 - 16 = 9

c = 3

e = \frac{c}{a} = \frac{3}{5} = \mathbf{0.6}

The foci lie on the major axis (x-axis) at $(\pm c, 0)$ :

\boxed{\text{Foci} = (\pm 3, 0), \quad e = \frac{3}{5}}

Why This Works

An ellipse has two special points — the foci — such that the sum of distances from any point on the ellipse to both foci is constant (equal to $2a$ ). The eccentricity $e = c/a$ measures how “stretched” the ellipse is. When $e = 0$ , it’s a circle; as $e \to 1$ , it flattens into a line segment.

The relation $c^2 = a^2 - b^2$ comes from the geometry: $a$ is the semi-major axis, $b$ is the semi-minor axis, and $c$ is the distance from centre to focus. By the Pythagorean-like relationship in the ellipse, these three are connected.

For our ellipse, $e = 3/5 = 0.6$ — moderately elongated. The foci at $(\pm 3, 0)$ are inside the ellipse, 3 units from the centre along the x-axis.

Alternative Method — Quick check using the ratio

If you remember that for an ellipse, $b^2 = a^2(1 - e^2)$ :

16 = 25(1 - e^2) \implies 1 - e^2 = \frac{16}{25} \implies e^2 = \frac{9}{25} \implies e = \frac{3}{5}

For CBSE boards, always state which axis is the major axis. If $a^2$ is under $x^2$ , major axis is x-axis and foci are at $(\pm c, 0)$ . If $a^2$ is under $y^2$ , major axis is y-axis and foci are at $(0, \pm c)$ . Getting this wrong flips your foci to the wrong axis — instant 2-mark loss.

Common Mistake

Students often confuse the formulas for ellipse and hyperbola. For an ellipse: $c^2 = a^2 - b^2$ (subtract). For a hyperbola: $c^2 = a^2 + b^2$ (add). Using the wrong formula here gives $c^2 = 25 + 16 = 41$ , leading to $e = \sqrt{41}/5 > 1$ — which should immediately signal an error, since an ellipse must have $0 < e < 1$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Quick check using the ratio

Common Mistake

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