Conic Sections: Exam-Pattern Drill (6)

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Tags Conic Sections

Question

The equation of an ellipse is x225+y29=1\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1. Find the eccentricity, foci, length of latus rectum, and the equation of directrices.

Solution — Step by Step

Comparing with x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1, we have a2=25a^2 = 25 and b2=9b^2 = 9. Since a2>b2a^2 > b^2, the major axis is along the x-axis. So a=5a = 5 and b=3b = 3.

For an ellipse, b2=a2(1−e2)b^2 = a^2(1 - e^2), so e=1−b2/a2=1−9/25=16/25=4/5e = \sqrt{1 - b^2/a^2} = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5.

c=ae=5×4/5=4c = ae = 5 \times 4/5 = 4.

Foci at (±c,0)=(±4,0)(\pm c, 0) = (\pm 4, 0).

Directrices: x=±a/e=±5/(4/5)=±25/4x = \pm a/e = \pm 5/(4/5) = \pm 25/4.

ℓ=2b2a=2×95=185=3.6\ell = \frac{2 b^2}{a} = \frac{2 \times 9}{5} = \frac{18}{5} = 3.6

Why This Works

The ellipse formulas all stem from the geometric definition: the locus of points whose sum of distances from two foci is constant (=2a= 2a). From this, b2=a2−c2b^2 = a^2 - c^2 and e=c/ae = c/a follow naturally.

The latus rectum is the chord through a focus perpendicular to the major axis. Its length comes from substituting x=cx = c in the ellipse equation, getting y=±b2/ay = \pm b^2/a, so length is 2b2/a2b^2/a.

Memorize this 4-row table for ellipse x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 (a>ba > b):

QuantityFormula
Eccentricitye=1−b2/a2e = \sqrt{1 - b^2/a^2}
Foci(±ae,0)(\pm ae, 0)
Directricesx=±a/ex = \pm a/e
Latus rectum2b2/a2b^2/a

For b>ab > a (vertical major axis), swap roles of xx and yy throughout.

Alternative Method

Use the focal chord property: r1+r2=2ar_1 + r_2 = 2a for any point on the ellipse. At the latus rectum endpoint above the right focus, r1=r2r_1 = r_2 (semi-latus-rectum), so r1=ar_1 = a. Then by the right triangle with legs b2/ab^2/a and cc, you can recover the geometry. Same final answers.

Students mix up aa and bb when the equation has the larger denominator under y2y^2. Always check: the larger of a2,b2a^2, b^2 corresponds to the major axis direction. If under y2y^2, the foci lie on the y-axis.

Final answer: e=4/5e = 4/5, foci (±4,0)(\pm 4, 0), directrices x=±25/4x = \pm 25/4, latus rectum =18/5= 18/5.

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