Lesson 7 of 12 · 7 min
Standard equations of an ellipse
NCERT §10.5.3
Meera wants the string garden's oval as an equation, so it can be drawn on the terrace plan. With the centre at the origin and the pegs at (±4, 0), what is it?
The lesson in notes
In short
Put the centre at the origin with the foci F₁(−c, 0) and F₂(c, 0). Writing PF₁ + PF₂ = 2a with the distance formula, isolating one root, squaring, simplifying and squaring again leads to x²/a² + y²/(a² − c²) = 1, that is x²/a² + y²/b² = 1.
Conversely, a point on x²/a² + y²/b² = 1 has PF₁ = a + (c/a)x and PF₂ = a − (c/a)x, which add to 2a, so the equation describes exactly the ellipse.
With the foci on the y-axis the equation is x²/b² + y²/a² = 1. These two are the standard equations of an ellipse; ellipses with other centres or tilted axes are outside this chapter.
From x²/a² ≤ 1, −a ≤ x ≤ a and similarly −b ≤ y ≤ b: the ellipse sits inside the rectangle formed by x = ±a and y = ±b and touches its four sides.
If (x, y) is on the ellipse, so are (−x, y), (x, −y) and (−x, −y): an ellipse is symmetric about both coordinate axes.
Both foci sit on the major axis, and the bigger of the two denominators tells which axis that is: under x² it is the x-axis, under y² the y-axis.
x²/25 + y²/9 = 1: a = 5, b = 3, c = √(25 − 9) = 4. Foci (±4, 0), vertices (±5, 0), major axis 10, minor axis 6, e = 4/5.
9x² + 4y² = 36 divided by 36 is x²/4 + y²/9 = 1. Now y² has the larger denominator, so a = 3, b = 2 and c = √5: foci (0, ±√5), vertices (0, ±3), major axis 6, minor axis 4, e = √5/3.