Conic Sections: Step-by-Step Worked Examples (4)

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Question

Find the equation of the ellipse whose foci are at (±4,0)(\pm 4, 0) and the length of the minor axis is 66. Also find the eccentricity and the directrices.

Solution — Step by Step

Foci on the x-axis means the major axis is horizontal. Standard form:

x2a2+y2b2=1,a>b\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b

with foci at (±c,0)(\pm c, 0) where c2=a2b2c^2 = a^2 - b^2.

Given c=4c = 4 and minor axis =2b=6= 2b = 6, so b=3b = 3.

a2=b2+c2=9+16=25    a=5a^2 = b^2 + c^2 = 9 + 16 = 25 \implies a = 5

x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 1

e=ca=45=0.8e = \frac{c}{a} = \frac{4}{5} = 0.8

Directrices for a horizontal ellipse: x=±a/ex = \pm a/e.

x=±54/5=±254x = \pm \frac{5}{4/5} = \pm \frac{25}{4}

Final: x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 1, e=4/5e = 4/5, directrices x=±25/4x = \pm 25/4.

Why This Works

The relation c2=a2b2c^2 = a^2 - b^2 comes directly from the geometric definition of an ellipse — the sum of distances from any point to the two foci equals 2a2a. Combined with the rule that minor axis = 2b2b and foci sit on the major axis at distance cc, every ellipse is fully determined by any two of {a,b,c}\{a, b, c\}.

Eccentricity measures how far the ellipse is from being circular. e=0e = 0 is a circle; e1e \to 1 is a flattened ellipse approaching a parabola.

Alternative Method

Use the focal-radius definition. The sum of distances from any point (x,y)(x, y) on the ellipse to the foci (±4,0)(\pm 4, 0) equals 2a2a. From the minor-axis endpoint (0,3)(0, 3):

16+9+16+9=2×5=10    2a=10\sqrt{16 + 9} + \sqrt{16 + 9} = 2 \times 5 = 10 \implies 2a = 10

So a=5a = 5 directly.

Common Mistake

Students mix up the ellipse and hyperbola relations. For an ellipse, c2=a2b2c^2 = a^2 - b^2 (with a>ba > b). For a hyperbola, c2=a2+b2c^2 = a^2 + b^2. Confusing these flips the answer’s sign and breaks the problem.

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