Mathematics · Ch 10 — Conic Sections

Relationship Between Semi-major Axis, Semi-minor Axis and the Distance of the Focus From the Centre of the Ellipse

10.5.1

Relationship Between Semi-major Axis, Semi-minor Axis and the Distance of the Focus From the Centre of the Ellipse

The Fundamental Relation of an Ellipse

Every ellipse is defined by three key numbers: the semi-major axis $a$ , the semi-minor axis $b$ , and the distance $c$ from the centre to each focus. These three are not independent — they are tied together by a single, elegant relationship that follows directly from the definition of the ellipse itself.

Recall the definition: an ellipse is the set of all points $P$ such that the sum of the distances to the two fixed foci $F_1$ and $F_2$ is constant. That constant is $2a$ , the length of the major axis. We will now use two special points on the ellipse — one at the end of the major axis and one at the end of the minor axis — to discover how $a$ , $b$ , and $c$ are connected.

Using a Point on the Major Axis

Consider the point $P$ at the right end of the major axis. Its coordinates are $(a, 0)$ . The foci are at $(-c, 0)$ and $(c, 0)$ .

The sum of the distances from $P$ to the two foci is:

F_1P + F_2P

From the geometry of the figure, $F_1P = F_1O + OP$ , where $O$ is the centre. Since $F_1O = c$ and $OP = a$ , we have $F_1P = c + a$ . The distance $F_2P$ is simply $a - c$ (the distance from the focus at $(c,0)$ to the point $(a,0)$ ).

Therefore:

F_1P + F_2P = (c + a) + (a - c) = 2a

This is exactly the constant sum we expect from the definition. No new relation yet — it simply confirms that the constant is $2a$ .

Using a Point on the Minor Axis

Now take the point $Q$ at the top end of the minor axis. Its coordinates are $(0, b)$ . The foci are still at $(-c, 0)$ and $(c, 0)$ .

The distance from $Q$ to the focus $F_1(-c, 0)$ is:

F_1Q = \sqrt{(0 + c)^2 + (b - 0)^2} = \sqrt{c^2 + b^2}

Similarly, the distance to $F_2(c, 0)$ is:

F_2Q = \sqrt{(0 - c)^2 + (b - 0)^2} = \sqrt{c^2 + b^2}

So the sum of the distances is:

F_1Q + F_2Q = \sqrt{c^2 + b^2} + \sqrt{c^2 + b^2} = 2\sqrt{c^2 + b^2}

Equating the Two Results

Since both $P$ and $Q$ lie on the same ellipse, the sum of the distances to the foci must be the same for both points. From the definition, that constant sum is $2a$ . Therefore:

2\sqrt{c^2 + b^2} = 2a

Divide both sides by 2:

\sqrt{c^2 + b^2} = a

Square both sides:

c^2 + b^2 = a^2

This is the fundamental relationship.

a^2 = b^2 + c^2

Rearranging, we can also write:

c = \sqrt{a^2 - b^2}

and

b = \sqrt{a^2 - c^2}

Important

This relation is the Pythagorean identity of the ellipse. It tells you that $a$ is always the largest of the three numbers — the hypotenuse of a right triangle whose legs are $b$ and $c$ .

What This Tells Us Geometrically …

Figure 10.23Relation among a, b, c

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

The figure shows a standard ellipse centred at $O$ , with its major axis horizontal. The two foci are labelled $F_1$ and $F_2$ , placed symmetrically on the major axis at distances $c$ from the centre. Two special points are marked: $P$ at the right vertex (the end of the major axis) and $Q$ at the top of the minor axis (the end of the minor axis). Dashed lines and arrows indicate the lengths $a$ (semi-major axis, from centre to vertex), $b$ (semi-minor axis, from centre to the top), $c$ (distance from centre to each focus), and $a-c$ (the distance from the right focus $F_2$ to the vertex $P$ ). Two blue line segments from $Q$ to $F_1$ and $Q$ to $F_2$ are drawn, each of length $\sqrt{b^2 + c^2}$ .

The physical idea is to use the definition of an ellipse — that the sum of distances from any point on the ellipse to the two foci is constant — and apply it to two convenient points: the vertex $P$ and the top of the minor axis $Q$ . By equating the two sums, we derive the fundamental relation among $a$ , $b$ , and $c$ .

For point $P$ at the right vertex, the distances to the foci are:

$F_1P = F_1O + OP = c + a$
$F_2P = a - c$ Adding them gives $(c + a) + (a - c) = 2a$ .

For point $Q$ at the top of the minor axis, the distances to the foci are equal by symmetry. Using the right triangle formed by $O$ , $Q$ , and either focus, each distance is $\sqrt{b^2 + c^2}$ . So the sum is $2\sqrt{b^2 + c^2}$ .

Since both $P$ and $Q$ lie on the same ellipse, these two sums must be equal:

2\sqrt{b^2 + c^2} = 2a

Cancelling the factor of 2 gives $\sqrt{b^2 + c^2} = a$ , and squaring both sides yields the central relation:

a^2 = b^2 + c^2

This is the Pythagorean relation for an ellipse. It tells us that $a$ is the hypotenuse of a right triangle whose legs are $b$ and $c$ . Rearranging, we also get $c = \sqrt{a^2 - b^2}$ , which is how the focal distance is computed from the semi-axes. …

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