This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1915 edition. Excerpt: ...1) (y--2) V--3, but the locus has one real point. Irreducible Equations.--It will be shown later (Art. 59) that the locus of an irreducible equation of the second degree is an ellipse, parabola, hyperbola or entirely imaginary. A circle is considered as an ellipse with axes of equal length. By the second degree part of an equation of the second degree is meant the part Ax2 + Bxy + Cy2 (346) containing the terms of second degree. We shall now show how to determine by inspection of this second degree part whether a given second degree equation represents an ellipse, parabola or hyperbola. An ellipse whose axes are the lines Aa + By + Ci = 0, A& + By + C2 = 0 is represented by the equation 1-/in + BiV + Ci 1 tAg + Bjy + C, y = 1 a2 VJJrB / VA22 + Bi I The second degree part of this equation is m (Ai? + fliy)2 (A2x + BiVy 1; o2 (Ai2 + Bi2) + 62 (A/ + 22). Since this is a sum of squares it has imaginary factors. If the axis of a parabola is Aix + Biy + & = (), and the line through the vertex perpendicular to the axis is A2x + Bty + C2 = 0, the equation of the curve is Mg + By + CA2 = /Aa + Bw + C, VAi2 + Bi2 / "I VAi + Bi I The second degree part of this equation is (A.x + Bgy W A? + B?' which is a complete square. which is a product of real first degree factors. Inspection of (1), (2) and (3) shows that the equations of ellipse, parabola and hyperbola are distinguished by the fact that the second degree part has imaginary factors in case of the ellipse, is a complete square in case of the parabola, and has real and distinct factors in case of the hyperbola. Example 1. Show that 8 x2-8 xy + 2 y2 = 2 x-3 is the equation of a parabola. Solving for y, Fig. 346. Solving for y, y = 2x $ Vix-6. Since y is an irrational...
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