Ellipse standard equation

The defining property. An ellipse is the locus of a point $P(x,y)$ such that the sum of its distances from two fixed points $F_1$ and $F_2$, the foci, stays constant. Call that constant sum $2a$, where $2a$ must be greater than the distance between the foci for the locus to be a real, closed curve rather than empty.

Placing coordinates conveniently. Put the centre of the ellipse at the origin and the foci on the $x$-axis at $F_1(-c,0)$ and $F_2(c,0)$, symmetric about the origin. This symmetry is a choice, not a law of nature — it is what turns the algebra into something manageable. For any point $P(x,y)$ on the locus:

where: $x, y$ are the coordinates of a general point on the ellipse, $c$ is the distance from the centre to each focus, and $2a$ is the constant sum of distances fixed by the string length.

From two square roots to one clean equation. Isolating one radical and squaring twice, as carried out in full in the derivation below, eliminates both square roots and leaves an equation in $x^2$ and $y^2$ alone. Writing $b^2 = a^2 - c^2$ (valid since $a>c$ for a real ellipse) turns that equation into the standard form

where: $a$ is the semi-major axis length, $b$ is the semi-minor axis length, and the constraint $a>b$ is a labelling convention — whichever axis is longer is always called $a$, regardless of whether it lies along $x$ or $y$.

Reading off the parts. For $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b$: the major axis lies along the $x$-axis with length $2a$ and vertices $(\pm a,0)$; the minor axis lies along the $y$-axis with length $2b$ and endpoints $(0,\pm b)$; and the foci sit at $(\pm c,0)$ with $c=\sqrt{a^2-b^2}$. If instead $b>a$, every role of $x$ and $y$ swaps: the major axis runs along the $y$-axis, vertices are $(0,\pm b)$, and foci are $(0,\pm c)$ with $c=\sqrt{b^2-a^2}$.

Problem. Find the equation of the ellipse with foci $(\pm 4,0)$ and vertices $(\pm 5,0)$, then verify the sum-of-distances property at a vertex.

Step 1. The vertices are at $(\pm a, 0)$, so $a = 5$ and $a^2 = 25$.

Step 2. The foci are at $(\pm c, 0)$, so $c = 4$.

Step 3. Using $b^2 = a^2 - c^2 = 25 - 16 = 9$.

Step 4. Substitute $a^2=25$ and $b^2=9$ into the standard form to get the equation of the ellipse: $\boxed{\dfrac{x^2}{25}+\dfrac{y^2}{9}=1}$.

Step 5 (verification). At the vertex $(5,0)$, the distance to focus $(4,0)$ is $1$ and the distance to focus $(-4,0)$ is $9$, and their sum is $1+9=10=2a$, confirming $a=5$ as required by the defining property.

Competitive stretch

Reverse the labelling: for $\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$, compare denominators — $25>16$, so the major axis is along the $y$-axis, giving $a^2=25$, $b^2=16$, $a=5$, $b=4$, and $c=\sqrt{a^2-b^2}=\sqrt{9}=3$. The foci are at $(0,\pm 3)$ and the vertices at $(0,\pm 5)$, with major axis length $2a=10$ and minor axis length $2b=8$ — a reminder that $a$ always names the semi-major axis, whichever coordinate it multiplies.

• Assuming the larger denominator always sits under $x^2$ — it does not; compare the two denominators and call the larger one $a^2$, whichever axis it belongs to.
• Confusing the ellipse relation $c^2=a^2-b^2$ with the hyperbola's $c^2=a^2+b^2$ — the minus sign here reflects that $c<a$ for a real ellipse.
• Treating $a>b>0$ as meaning "$x^2$'s denominator is bigger" rather than a naming convention for whichever axis is longer.
• Forgetting to take the positive square root when solving $c=\sqrt{a^2-b^2}$, or mixing up which of $a$, $b$, $c$ was given in a problem.

Ask the tutor: want to see how this same squaring method changes when the foci are placed on the y-axis instead, or how eccentricity connects back to $a$, $b$, $c$?