The Ellipse

Definition, the a-b-c Triangle, and Eccentricity

Key Point (Definition): An ellipse is the set of all points whose distances from two fixed points (the foci $F_1$, $F_2$) have a constant sum — always greater than the distance between the foci.

The midpoint of $F_1F_2$ is the centre; the line through the foci is the major axis (length $2a$); the perpendicular line through the centre is the minor axis (length $2b$); the ends of the major axis are the vertices. Write $2c$ for the distance between the foci.

Where $c^2 = a^2 - b^2$ comes from (the two-point argument): the vertex end of the major axis gives sum $= (c + a) + (a - c) = 2a$; the end of the minor axis gives sum $= 2\sqrt{b^2 + c^2}$. Setting them equal:

$$a = \sqrt{b^2 + c^2}, \quad\text{i.e.}\quad c^2 = a^2 - b^2$$

Key Point (Eccentricity): $e = \frac{c}{a}$, with $0 < e < 1$ for a genuine ellipse. The foci sit at distance $ae$ from the centre. As $e \to 0$ the foci merge and the ellipse becomes a circle ($b \to a$).

Deriving the standard equation. With centre at the origin and foci $(\pm c, 0)$, the condition $PF_1 + PF_2 = 2a$ becomes, after two squarings,

$$\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1 \;\Rightarrow\; \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

Along the way one finds the tidy focal distances $PF_1 = a + \frac{c}{a}x$ and $PF_2 = a - \frac{c}{a}x$ — worth remembering for the JEE Corner. The curve lives in the box $-a \le x \le a$, $-b \le y \le b$ and is symmetric about both axes.

Both Orientations and the Latus Rectum

Key Point (Standard equations): Major axis on the $x$-axis: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with foci $(\pm c, 0)$. Major axis on the $y$-axis: $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ with foci $(0, \pm c)$. In both, $a > b$ and $c^2 = a^2 - b^2$.

Which is which? The larger denominator sits under the major-axis variable, and the foci always lie on the major axis.

Key Point (Latus rectum): the focal chord perpendicular to the major axis has length $\frac{2b^2}{a}$ — substitute $x = ae$ into the equation and the semi-length comes out as $\frac{b^2}{a}$.

The standard workflow:

Given the equation, read $a$ and $b$ from the denominators, compute $c = \sqrt{a^2 - b^2}$, then list foci, vertices, axis lengths, $e = \frac{c}{a}$ and latus rectum $\frac{2b^2}{a}$. For $\frac{x^2}{25} + \frac{y^2}{9} = 1$: $c = 4$, foci $(\pm 4, 0)$, vertices $(\pm 5, 0)$, $e = \frac{4}{5}$, latus rectum $\frac{18}{5}$.

Given data, pick the orientation from where the foci/vertices lie, then solve for the missing constant: vertices $(\pm 13, 0)$ and foci $(\pm 5, 0)$ give $b^2 = 169 - 25 = 144$, so $\frac{x^2}{169} + \frac{y^2}{144} = 1$. Two points on the ellipse give two linear equations in $\frac{1}{a^2}$ and $\frac{1}{b^2}$.

[Board Tip] $c^2 = a^2 - b^2$ for the ellipse but $c^2 = a^2 + b^2$ for the hyperbola. Mixing the two is the single most common conics error — anchor it as "ellipse: $a$ is the hypotenuse."

Solved Examples

Example 1: Full anatomy from the equation

For $\frac{x^2}{25} + \frac{y^2}{9} = 1$, find the foci, vertices, axis lengths, eccentricity and latus rectum.

Solution:

Step 1 — Read $a$ and $b$. Larger denominator under $x^2$: major on the $x$-axis, $a = 5$, $b = 3$.

Step 2 — Compute $c$. $c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = 4$.

Step 3 — List everything. Foci $(\pm 4, 0)$; vertices $(\pm 5, 0)$; major axis 10; minor axis 6; $e = \frac{c}{a} = \frac{4}{5}$; latus rectum $\frac{2b^2}{a} = \frac{18}{5}$.

Takeaway: One $c$-computation unlocks the entire list — everything else is direct substitution.

Example 2: Major axis on the $y$-axis

Analyse $9x^2 + 4y^2 = 36$.

Solution:

Step 1 — Normalise. Divide by 36: $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

Step 2 — Orient. The larger denominator (9) sits under $y^2$: major on the $y$-axis, $a = 3$, $b = 2$.

Step 3 — Compute and list. $c = \sqrt{9 - 4} = \sqrt{5}$: foci $(0, \pm\sqrt{5})$, vertices $(0, \pm 3)$, major axis 6, minor axis 4, $e = \frac{\sqrt{5}}{3}$, latus rectum $\frac{2(4)}{3} = \frac{8}{3}$.

Takeaway: Always divide to "$= 1$" form first — the raw coefficients invert the apparent roles of $a$ and $b$.

Example 3: Reading practice

Find foci, eccentricity and latus rectum of: (i) $\frac{x^2}{36} + \frac{y^2}{16} = 1$; (ii) $\frac{x^2}{4} + \frac{y^2}{25} = 1$; (iii) $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

Solution:

Step 1 — (i). $a = 6, b = 4$: $c = \sqrt{20} = 2\sqrt{5}$: foci $(\pm 2\sqrt{5}, 0)$, $e = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$, latus rectum $\frac{2(16)}{6} = \frac{16}{3}$.

Step 2 — (ii). Major on $y$ ($25 > 4$): $a = 5, b = 2$, $c = \sqrt{21}$: foci $(0, \pm\sqrt{21})$, $e = \frac{\sqrt{21}}{5}$, latus rectum $\frac{8}{5}$.

Step 3 — (iii). $a = 4, b = 3$, $c = \sqrt{7}$: foci $(\pm\sqrt{7}, 0)$, $e = \frac{\sqrt{7}}{4}$, latus rectum $\frac{2(9)}{4} = \frac{9}{2}$.

Takeaway: Foci follow the major axis — check which denominator wins before writing any coordinates.

Example 4: Rescale first

Analyse: (i) $36x^2 + 4y^2 = 144$; (ii) $16x^2 + y^2 = 16$; (iii) $4x^2 + 9y^2 = 36$.

Solution:

Step 1 — (i). $\frac{x^2}{4} + \frac{y^2}{36} = 1$: $a = 6$, $b = 2$, $c = \sqrt{32} = 4\sqrt{2}$; foci $(0, \pm 4\sqrt{2})$, $e = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}$, latus rectum $\frac{2(4)}{6} = \frac{4}{3}$.

Step 2 — (ii). $x^2 + \frac{y^2}{16} = 1$: $a = 4$, $b = 1$, $c = \sqrt{15}$; foci $(0, \pm\sqrt{15})$, $e = \frac{\sqrt{15}}{4}$, latus rectum $\frac{2(1)}{4} = \frac{1}{2}$.

Step 3 — (iii). $\frac{x^2}{9} + \frac{y^2}{4} = 1$: $a = 3$, $b = 2$, $c = \sqrt{5}$; foci $(\pm\sqrt{5}, 0)$, $e = \frac{\sqrt{5}}{3}$, latus rectum $\frac{8}{3}$.

Takeaway: Divide by the constant, THEN compare denominators — the analysis is mechanical after that.

Example 5: From vertices and foci

Find the ellipse with: (i) vertices $(\pm 13, 0)$, foci $(\pm 5, 0)$; (ii) vertices $(\pm 5, 0)$, foci $(\pm 4, 0)$; (iii) vertices $(0, \pm 13)$, foci $(0, \pm 5)$; (iv) vertices $(\pm 6, 0)$, foci $(\pm 4, 0)$.

Solution:

Step 1 — (i). $a = 13$, $c = 5$: $b^2 = 169 - 25 = 144$: $\frac{x^2}{169} + \frac{y^2}{144} = 1$.

Step 2 — (ii). $b^2 = 25 - 16 = 9$: $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Step 3 — (iii). Same numbers on the $y$-axis: $\frac{x^2}{144} + \frac{y^2}{169} = 1$.

Step 4 — (iv). $b^2 = 36 - 16 = 20$: $\frac{x^2}{36} + \frac{y^2}{20} = 1$.

Takeaway: Vertices give $a$, foci give $c$, and $b^2 = a^2 - c^2$ finishes — orientation just swaps the denominators.

Example 6: From an axis length and foci

Find the ellipse with: (i) major axis 20, foci $(0, \pm 5)$; (ii) major axis 26, foci $(\pm 5, 0)$; (iii) minor axis 16, foci $(0, \pm 6)$.

Solution:

Step 1 — (i). $2a = 20$: $a = 10$, $c = 5$: $b^2 = 100 - 25 = 75$: $\frac{x^2}{75} + \frac{y^2}{100} = 1$.

Step 2 — (ii). $a = 13$, $c = 5$: $b^2 = 144$: $\frac{x^2}{169} + \frac{y^2}{144} = 1$.

Step 3 — (iii). MINOR axis 16 gives $b = 8$; $c = 6$: $a^2 = b^2 + c^2 = 100$: $\frac{x^2}{64} + \frac{y^2}{100} = 1$.

Takeaway: Given $b$ and $c$, the missing $a^2$ comes from $a^2 = b^2 + c^2$ — the relation rearranged, not a new formula.

Example 7: From the ends of the axes

Find the ellipse with: (i) ends of major axis $(\pm 3, 0)$ and ends of minor axis $(0, \pm 2)$; (ii) ends of major axis $(0, \pm\sqrt{5})$ and ends of minor axis $(\pm 1, 0)$.

Solution:

Step 1 — (i). $a = 3$ on $x$, $b = 2$: $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

Step 2 — (ii). $a = \sqrt{5}$ on $y$, $b = 1$: $x^2 + \frac{y^2}{5} = 1$.

Takeaway: Axis ends hand you $a$ and $b$ directly — no $c$ needed at all.

Example 8: From $a$, $b$, $c$ data

Find the ellipse with: (i) foci $(\pm 3, 0)$ and $a = 4$; (ii) $b = 3$, $c = 4$, centre at the origin, foci on the $x$-axis.

Solution:

Step 1 — (i). $c = 3$, $a = 4$: $b^2 = 16 - 9 = 7$: $\frac{x^2}{16} + \frac{y^2}{7} = 1$.

Step 2 — (ii). $a^2 = b^2 + c^2 = 9 + 16 = 25$: $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Takeaway: Two of $\{a, b, c\}$ always determine the third — identify which two you hold before anything else.

Example 9: Through two points

Find the ellipse with major axis along the $x$-axis passing through $(4, 3)$ and $(-1, 4)$.

Solution:

Step 1 — Set up linear equations. Let $u = \frac{1}{a^2}$, $v = \frac{1}{b^2}$: $16u + 9v = 1$ and $u + 16v = 1$.

Step 2 — Solve. Subtract $16\times$ the second from the first: $9v - 256v = 1 - 16$ gives $v = \frac{15}{247}$, then $u = 1 - 16v = \frac{7}{247}$.

Step 3 — Write the ellipse. $a^2 = \frac{247}{7}$, $b^2 = \frac{247}{15}$: $\frac{7x^2}{247} + \frac{15y^2}{247} = 1$, i.e. $7x^2 + 15y^2 = 247$.

Takeaway: The substitution $u = \frac{1}{a^2}, v = \frac{1}{b^2}$ turns two points into a 2×2 LINEAR system.

Example 10: Two points, both orientations

(i) Major axis on the $y$-axis, through $(3, 2)$ and $(1, 6)$. (ii) Major axis on the $x$-axis, through $(4, 3)$ and $(6, 2)$.

Solution:

Step 1 — (i) Linear system. $\frac{9}{b^2} + \frac{4}{a^2} = 1$ and $\frac{1}{b^2} + \frac{36}{a^2} = 1$: subtracting, $\frac{8}{b^2} = \frac{32}{a^2}$, so $a^2 = 4b^2$.

Step 2 — (i) Back-substitute. $\frac{9}{b^2} + \frac{1}{b^2} = 1$ gives $b^2 = 10$, $a^2 = 40$: $\frac{x^2}{10} + \frac{y^2}{40} = 1$.

Step 3 — (ii) Same method. $\frac{16}{a^2} + \frac{9}{b^2} = 1$ and $\frac{36}{a^2} + \frac{4}{b^2} = 1$ solve to $a^2 = 52$, $b^2 = 13$: $\frac{x^2}{52} + \frac{y^2}{13} = 1$.

Step 4 — Check (ii). $\frac{16}{52} + \frac{9}{13} = \frac{4}{13} + \frac{9}{13} = 1$ ✓.

Takeaway: Check one point at the end — the fractions make sign slips easy and the check makes them visible.