Ellipse
In short: An ellipse is a closed oval traced by a point that keeps the sum of its distances from two anchor points, the foci, unchanged; its centered equation is x²/a² + y²/b² = 1.
PrecalculusGeometryG-GPE.A.3
What is Ellipse?
An ellipse is a stretched circle. Pin the two ends of a string to a board, pull the string tight with a pencil, and slide the pencil around: the oval it draws is an ellipse, and the pins mark its foci.
Centered at the origin, \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\) with \(a > b > 0\) has a vertical major axis of length \(2a\) and a minor axis of length \(2b\). The foci lie on the major axis at distance \(c\) from the center, where \(c^2 = a^2 – b^2\). If the larger denominator sits under \(x^2\), the major axis is horizontal.
How to spot it
- An equation has x² and y² added together with different positive denominators, and equals 1.
- A problem mentions two fixed points whose distance sum stays constant, such as a whispering gallery.
- A planet’s orbit or a racetrack is modeled by an oval with a long and a short axis.
- You are asked for vertices, co-vertices or foci of a curve given by a sum of squares.
The method
- Scale the equation until its right side is exactly 1.
- Compare the two denominators: the bigger one is \(a^2\), and its variable names the long direction.
- Compute \(c = \sqrt{a^2 – b^2}\) to locate the foci on the major axis.
- Mark the vertices at distance \(a\), the co-vertices at distance \(b\), and sketch the oval through them.
Ellipse: solved example
Problem
For \(25x^2 + 9y^2 = 225\), find the lengths of the axes, the vertices and the foci.
Solution
Dividing everything by 225 gives \(\frac{x^2}{9} + \frac{y^2}{25} = 1\). Since 25 sits under \(y^2\), \(a = 5\), \(b = 3\), and the long direction is up and down.
Then \(c = \sqrt{25 – 9} = 4\). Lengths: 10 along the long axis, 6 along the short one. Vertices \((0, \pm 5)\), co-vertices \((\pm 3, 0)\), and foci \((0, \pm 4)\).
Try it
Question 1
Find the foci of \(\frac{x^2}{49} + \frac{y^2}{4} = 1\).
Show the answer
\(a^2 = 49\), \(b^2 = 4\), so \(c = \sqrt{45} = 3\sqrt{5} \approx 6.708\); the foci are \((\pm 3\sqrt{5}, 0)\).
Question 2
An oval is centered at \((3, -2)\), reaching 5 units left and right and 3 units up and down. Write its equation.
Show the answer
\(\frac{(x – 3)^2}{25} + \frac{(y + 2)^2}{9} = 1\).
Ellipse vs. Hyperbola
Both use two foci and squared terms over denominators. For the oval, the squares are added and the curve is one closed loop; for the hyperbola, they are subtracted and the curve has two open branches. Also, \(c^2 = a^2 – b^2\) here, while the other uses \(c^2 = a^2 + b^2\).
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