For an ellipse centered at the origin the x intercepts are given by a 0 and a

# For an ellipse centered at the origin the x

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For an ellipse centered at the origin, the x-intercepts aregiven by (a,0) and (a,0),and the y-intercepts are givenby (0,b) and (0,b).A hyperbolais the set of all points (x,y) such that thedifference of the distances between (x,y) and two dis-tinct points is a constant.The fixed points are called thefoci of the hyperbola.x2a2y2b21ExamplesExample 1Example 2x225y291yx(a, 0)(0, 0)(0, b)(0, b)(a, 0)4545631231234561456312y6x2(5, 0)(0, 3)(5, 0)(0, 3)The standard form of an equation of a parabola withvertex (h,k) and horizontal axis of symmetry is xa(yk)2hThe equation of the axis of symmetry is yk.• If a0, then the parabola opens to the right.• If a0, the parabola opens to the left.Example 2Given the parabola ,determine the coordinates of the vertex and the equa-tion of the axis of symmetry.The vertex is (1, 0).The axis of symmetry is y0.23415yx1345213452345120(1, 0) x141y0221x14y21
Summary833Standard Forms of an Equation of a HyperbolaLet aand brepresent positive real numbers.Horizontal Transverse Axis.The standard form of anequation of a hyperbola with a horizontal transverseaxis and center at the origin is given byVertical Transverse Axis.The standard form of an equa-tion of a hyperbola with a vertical transverse axis andcenter at the origin is given byy2a2x2b21x2a2y2b21Example 3Example 4y24x2161x24y2161454521345312345121yx32(2, 0)(2, 0)454521345312345121yx32(0, 2)(0, 2)Nonlinear Systems of Equations in Two VariablesSection 11.4Key ConceptsA nonlinear system of equationscan be solved by graph-ing or by the substitution method.yx3, 3()3, 3( )x2y02x2y215ExamplesExample 1Solve for y:Substitute in first equation.If then If then Points of intersection are and 113, 32.113, 32y113223.x13y113223.x13x13x25orx23x250orx2301x2521x2320x42x21502x2x4152x21x22215Ayx2x2y0B2x2y215A
834Chapter 11Conic SectionsNonlinear Inequalities and Systems of InequalitiesSection 11.5Key ConceptsGraph a nonlinear inequality by using the test pointmethod.That is,graph the related equation.Then choosetest points in each region to determine where the in-equality is true.ExamplesExample 1Example 24513yx4531154345312222x2y2644513y4531154232245321yx2A nonlinear system may also be solved by using theaddition methodwhen the equations share liketerms.Example 2The points of intersection are 11, 122, and 11, 122.11, 122, 11, 122,y12y22If x1,21122y24y12y22If x1,21122y24x1x217x27777x273x25y2133x25y21310x25y2202x2y24Mult. by 5.
Review Exercises835Graph a system of nonlinear inequalities by finding theintersection of the solution sets. That is, graph the solu-tion set for each individual inequality, then take the in-tersection.
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