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Concept 10 / 14Coordinate Geometry

The Ellipse: Definitions, Eccentricity, Foci, Directrices and Focal Distances

The focus-directrix and two-foci definitions, the standard equation with b^2 = a^2(1 - e^2), axes, foci, directrices and latus rectum, focal distances a - ex and a + ex, parametric form and the auxiliary circle, and ellipses with major axis along y.

Figure 1
An ellipse is the locus of a point whose distance from a fixed point (a focus) is ee times its distance from a fixed line (the directrix), with 0<e<10 < e < 1. Equivalently, and more usefully, it is the locus of a point whose distances from two fixed points (the foci) have a constant sum, SP+S′P=2aSP + S'P = 2a. This is the 'pin and string' construction: loop a string of length 2a2a around two pins and trace with a pencil. Standard form: with foci (±ae,0)(\pm ae, 0), x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where b2=a2(1−e2)b^{2} = a^{2}(1 - e^{2}) and a>ba > b. Elements, shown in the figure: Centre (0,0)(0, 0); vertices (±a,0)(\pm a, 0); major axis length 2a2a along the xx-axis; minor axis length 2b2b. Foci (±ae,0)(\pm ae, 0), with ae=a2−b2ae = \sqrt{a^{2} - b^{2}}; eccentricity e=1−b2a2e = \sqrt{1 - \frac{b^{2}}{a^{2}}}. Directrices x=±aex = \pm\frac{a}{e}, outside the ellipse. Latus rectum length 2b2a\frac{2b^{2}}{a}, with ends (±ae,±b2a)(\pm ae, \pm\frac{b^{2}}{a}). The eccentricity measures flatness: e→0e \to 0 gives a circle (b=ab = a), and e→1e \to 1 gives a very elongated ellipse. If b>ab > a in x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, the major axis lies along the yy-axis: foci (0,±be)(0, \pm be) with a2=b2(1−e2)a^{2} = b^{2}(1 - e^{2}), and directrices y=±bey = \pm\frac{b}{e}. Always identify the larger denominator first. Focal distances. For P(x,y)P(x, y) on the ellipse, SP=a−exSP = a - ex and S′P=a+exS'P = a + ex (from the focus-directrix definition with each directrix). Their sum is 2a2a, confirming the two-foci definition. Parametric form and the auxiliary circle. The circle x2+y2=a2x^{2} + y^{2} = a^{2} on the major axis as diameter is the auxiliary circle. Every point of the ellipse can be written (acos⁡θ,bsin⁡θ)(a\cos\theta, b\sin\theta), where θ\theta is the eccentric angle: the point (acos⁡θ,asin⁡θ)(a\cos\theta, a\sin\theta) on the auxiliary circle, projected vertically onto the ellipse, gives the ellipse point. So the ellipse is the auxiliary circle squashed vertically by the factor ba\frac{b}{a}, which explains its area πab\pi ab (the circle's area πa2\pi a^{2} scaled by ba\frac{b}{a}). Shifted ellipses: (x−h)2a2+(y−k)2b2=1\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1 has centre (h,k)(h, k); all elements shift accordingly. Complete the square in both variables to reach this form. Position of a point: (x1,y1)(x_{1}, y_{1}) is inside, on or outside the ellipse according as x12a2+y12b2−1\frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}} - 1 is negative, zero or positive. Key Takeaways: - x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a>ba > b: b2=a2(1−e2)b^{2} = a^{2}(1 - e^{2}), foci (±ae,0)(\pm ae, 0), directrices x=±aex = \pm\frac{a}{e}. - SP+S′P=2aSP + S'P = 2a; SP=a−exSP = a - ex, S′P=a+exS'P = a + ex. - Latus rectum 2b2a\frac{2b^{2}}{a}. - Parametric (acos⁡θ,bsin⁡θ)(a\cos\theta, b\sin\theta); auxiliary circle x2+y2=a2x^{2} + y^{2} = a^{2}; area πab\pi ab. - If b>ab > a the major axis is vertical: foci (0,±be)(0, \pm be). - Distance between foci 2ae2ae; between directrices 2ae\frac{2a}{e}. - Pitfall: using b2=a2(e2−1)b^{2} = a^{2}(e^{2} - 1) (the hyperbola relation) for an ellipse. Brain-Teaser 1: Find the eccentricity, foci, directrices and latus rectum of 9x2+25y2=2259x^{2} + 25y^{2} = 225. Solution: Divide by 225225: x225+y29=1\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1, so a=5a = 5, b=3b = 3. e=1−925=45e = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}. Foci (±4,0)(\pm 4, 0); directrices x=±254x = \pm\frac{25}{4}; latus rectum 2×95=185\frac{2 \times 9}{5} = \frac{18}{5}. The 33-44-55 triangle formed by the centre, a focus and an end of the minor axis is a good memory aid: b2+(ae)2=a2b^{2} + (ae)^{2} = a^{2}. Brain-Teaser 2: Find the ellipse with foci (±2,0)(\pm 2, 0) passing through (2,3)(2, 3). Solution: ae=2ae = 2. The sum of focal distances from (2,3)(2, 3) is 3+16+9=3+5=8=2a3 + \sqrt{16 + 9} = 3 + 5 = 8 = 2a, so a=4a = 4 and b2=16−4=12b^{2} = 16 - 4 = 12. The ellipse is x216+y212=1\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1. Using the two-foci definition directly avoids solving for aa and bb from a pair of equations. Brain-Teaser 3: If the latus rectum of an ellipse equals half its minor axis, find the eccentricity. Solution: 2b2a=b\frac{2b^{2}}{a} = b, so b=a2b = \frac{a}{2}. Then e2=1−b2a2=1−14=34e^{2} = 1 - \frac{b^{2}}{a^{2}} = 1 - \frac{1}{4} = \frac{3}{4} and e=32e = \frac{\sqrt{3}}{2}. Relations between the lengths of axes and latus rectum are always converted into a ratio ba\frac{b}{a}, and then into ee.

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The Ellipse: Definitions, Eccentricity, Foci, Directrices and Focal Distances — DPP

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