Chapter 1 Topics in Analytic Geometry

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 9The derivation of these equations are similar to those already given for parabolas andellipses, so we will leave them as exercises.A Quick Way to Find AsymptotesThey are a trick that can be used find the equations of the asymptotes of a hyperbola.They can be obtained, when needed, by replacing 1 by 0 on the right side of the hyperbolaequation, and then solving for y in terms of x. For example, for the hyperbolawe would writex 2x 2a 2 − y2b 2 = 1a − y22 b = 0 of 2 y2 = b2which are the equations for the asymptotes.a 2 x2or y = ± b a xExample 1.5 Sketch the graph of the hyperbola9x 2 −4y 2 = 36,and showing its vertices, foci, and asymptotes.Solution .........Example 1.6 Find the equation of the hyperbola with vertices (0,±8) and asymptotes y =± 4 3 x.Solution .........Translated ConicsEquations of conic that are translated from their standard positions can be obtained byreplacing x by x − h and y by y − k in their standard equations. For a parabola, thistranslates the vertex from the origin to the point (h,k); and for ellipses and hyperbolas,this translates the center from the origin to the point (h,k).Parabolas with vertex (h,k) and axis parallel to x-axis(y −k) 2 = 4p(x−h) [Opens right](y −k) 2 = −4p(x−h) [Opens left]Parabolas with vertex (h,k) and axis parallel to y-axis(x−h) 2 = 4p(y −k)(x−h) 2 = −4p(y −k)[Opens up][Opens down]