Chapter 1 Topics in Analytic Geometry
Chapter 1 Topics in Analytic Geometry Chapter 1 Topics in Analytic Geometry
•MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 4OrientationyStandard Equation•(−p,0)xx = py 2 = −4pxy(0,p) •xy = −px 2 = 4pyyy = px•(0,−p)x 2 = −4pyTo illustrate how the equations in the above Figure are obtained, we will derive theequation for the parabola with focus (p,0) and directrix x = −p. Let P(x,y) be any pointon the parabola.yD(−p,y)P(x,y)•F(p,0)xx = −pSince P is equidistant from the focus and directrix, the distances PF and PD in the aboveFigure are equal; that is,PF = PD (1.1)From the distance formula, the distance PF and PD arePF = √ (x−p) 2 +y 2 and PD = √ (x+p) 2 (1.2)
•••••MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 5Substituting in (3.22) and squaring yieldsand after simplifying(x−p) 2 +y 2 = (x+p) 2 (1.3)y 2 = 4px (1.4)The derivations of the other equations in the previous Figure are similar.Example 1.1 Find the focus and directrix of the parabola x 2 = −6y and sketch its graph.Solution .........Example 1.2(a) Find an equation of a parabola that has vertex at the origin, opens right, and passesthrough the point P(7,−3).(b) Find the focus.Solution .........Equations of Ellipses in Standard PositionIt is traditional in the study of ellipse to denote the length of the major axis by 2a, thelength of the minor axis by 2b, and the distance between the foci by 2c.ccbbaaThe number a is called the semimajor axis and the number b the semiminor axis.There is a basic relationship between a, b, and c that can be obtained by examining thesum of the distances to the foci from a point P at the end of the major axis and from apoint Q at the end of the minor axis.√b2 +c 2Qbc c a−c√b2 +c 2•P
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•MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 4OrientationyStandard Equation•(−p,0)xx = py 2 = −4pxy(0,p) •xy = −px 2 = 4pyyy = px•(0,−p)x 2 = −4pyTo illustrate how the equations <strong>in</strong> the above Figure are obta<strong>in</strong>ed, we will derive theequation for the parabola with focus (p,0) and directrix x = −p. Let P(x,y) be any po<strong>in</strong>ton the parabola.yD(−p,y)P(x,y)•F(p,0)xx = −pS<strong>in</strong>ce P is equidistant from the focus and directrix, the distances PF and PD <strong>in</strong> the aboveFigure are equal; that is,PF = PD (1.1)From the distance formula, the distance PF and PD arePF = √ (x−p) 2 +y 2 and PD = √ (x+p) 2 (1.2)
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