Conic Sections: Common Mistakes and Fixes (1)

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Question

Find the equation of the ellipse whose foci are (±3,0)(\pm 3, 0) and the length of the latus rectum is 32/532/5.

Solution — Step by Step

Foci on the xx-axis means the major axis is horizontal:

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

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

c=3c = 3. Latus rectum =2b2/a=32/5= 2b^2/a = 32/5, so b2=16a/5b^2 = 16a/5.

9=a216a59 = a^2 - \tfrac{16a}{5} 5a216a45=05a^2 - 16a - 45 = 0 a=16±256+90010=16±115610=16±3410a = \tfrac{16 \pm \sqrt{256 + 900}}{10} = \tfrac{16 \pm \sqrt{1156}}{10} = \tfrac{16 \pm 34}{10}

So a=5a = 5 or a=1.8a = -1.8. Since a>0a > 0, a=5a = 5.

b2=16×55=16b^2 = \tfrac{16 \times 5}{5} = 16 x225+y216=1\boxed{\tfrac{x^2}{25} + \tfrac{y^2}{16} = 1}

Why This Works

Two facts pin down an ellipse with centre at origin and axes along coordinate axes: cc (focal distance) and one of aa, bb or the latus rectum. We had cc and the latus rectum, and the fundamental identity c2=a2b2c^2 = a^2 - b^2 closed the loop.

The latus rectum formula 2b2/a2b^2/a comes from substituting x=cx = c in the ellipse equation: y=±b2/ay = \pm b^2/a, giving full chord length 2b2/a2b^2/a.

Alternative Method

We could use eccentricity e=c/ae = c/a. Then b2=a2(1e2)b^2 = a^2(1-e^2) and latus rectum =2b2/a=2a(1e2)= 2b^2/a = 2a(1-e^2). Two equations in two unknowns (a,ea, e). Same answer through different algebra.

Common Mistake

The most common error is assuming a<ba < b for an ellipse with foci on the xx-axis. The major axis (and therefore aa) is along whichever axis the foci lie on. If the foci are on yy-axis, the form becomes x2b2+y2a2=1\tfrac{x^2}{b^2} + \tfrac{y^2}{a^2} = 1 with a>ba > b and foci (0,±c)(0, \pm c).

Quick latus rectum check: for 25x2+16y2=40025x^2 + 16y^2 = 400, divide by 400400: x216+y225=1\tfrac{x^2}{16} + \tfrac{y^2}{25} = 1. Here a2=25a^2 = 25 (foci on yy-axis), latus rectum =2(16)/5=32/5= 2(16)/5 = 32/5. Practise spotting which axis is major in 55 seconds.

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