2 Ellipse: the set of all points P in a plane such that the sum of the distances from P to two fixed points is a constant.(focus pl. foci)Standard Equation (center at (0, 0))Foci on x-axis Foci on y-axisMajor axis = 2aMinor axis = 2b(0, a)(0, b)minor axismajor(a, 0)(–b, 0)(b, 0)(–a, 0)major axisminor(0, –b)(0, –a)Foci:c2 = a2 – b2Foci: (±c, 0)Foci: (0, ±c)
3 Ex 1) Determine the endpoints of the major and minor axes and the foci of the ellipse 25x2 + 9y2 = 225 and graph.Standard form:c2 = 25 – 9c2 = 16c = 4 on y-axis(0, ±4)b2a2(±3, 0)(0, ±5)The shape of an ellipse is measured by a constant called the eccentricity.For an ellipse, e is equal to the ratio of the distance between the center and a focus to the distance between the center and a corresponding vertex < e < 1 iff the conic is an ellipse
4 Ex 2) The dwarf planet Pluto has an elliptical orbit with the sun at one focus. The minimum & maximum distances of Pluto from the sun occur at the vertices of the ellipse. The minimum distance is 2.7 billion miles and the maximum distance is 4.5 billion miles. Find the eccentricity of Pluto’s orbit.major axis = 2a2a = max dist + min distance2a = = 7.2a = 3.6c = 3.6 – 2.7 = 0.94.52.7
5 What if center is not the origin? Standard form of ellipse with center (h, k) & with axes parallel tocoordinate axisOREx 3) Determine an equation of an ellipse in standard form with foci(8, 3) and (–4, 3) if length of major axis is 14. Graph it.(Hint: graph as you go)center halfway between foci½(–4 + 8) = 2C(2, 3) & c = 6major axis = 14 a = 7 on “x-axis”c2 = a2 – b236 = 49 – b2b2 = 13b ≈ 3.6
6 The general form of an equation of an ellipse with axes parallel to the coordinate axes is Ax2 + Cy2 + Dx + Ey + F = 0.You can change from general form to standard form by …completing the square! Ex 4) Determine the center, the endpoints of the major & minor axes, the foci, and the eccentricity of x2 + 4y2 – 6x – 16y – 11 = 0. Graph it.(x2 – 6x ) + 4(y2 – 4y ) =(x – 3)2 + 4(y – 2)2 = 363636a2b2Center (3, 2)
7 6Major axis: (3 ± a, 2) (9, 2) and (–3, 2)Minor axis: (3, 2 ± b) (3, 5) and (3, –1)c2 = 36 – 9c2 = 273a2b2Center (3, 2)
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