Math 1202: Pre-Calculus Unit 2 Study Guide

Math 1208: Pre-Calculus Unit 8 Study Guide1. Be able to find the equation of a circle in standard form. The standard equation ofa circle is (x – h) 2 + (y – k) 2 = r 2 , where h is the x-value of the center, k is the y-value of the center, and r is the radius.2. Know how to find the center and radius from the equation of a circle. From thestandard equation of a circle, (h, k) is the center and r is the radius.3. Be able to find the equation of a circle in general form. The general form of acircle is x 2 + y 2 + dx + ey + f = 0.4. Know how to change the equation of a circle from standard form to general form:a. Multiply the binomial squares to get rid of the parentheses.b. Simplify.5. Know how to change the equation of a circle from general form to standard form:a. Group like terms.b. Complete the squares.c. Simplify.6. Know how to find the domain and the range of a circle.a. The domain is the x-value of the center ± the radius and all of the values inbetween.b. The range is the y-value of the center ± the radius and all of the values inbetween.7. Be able to find the equation of a circle when given three points on the circle.a. Use the general equation of a circle: x 2 + y 2 + dx + ey + f = 0.b. Put each point into the general equation for x and y. You should get 3different equations with d, e, and f.c. Using the substitution method or the opposite-coefficients method, solvefor d, e, and f.d. Substitute these numbers for d, e, and f in the general equation.8. Know how to find the midpoint of a line segment when given the endpoints. Themidpoint of a line segment connecting point (x 1 , y 1 ) with point (x 2 , y 2 ) is⎛ x1 + x2y1+ y2⎞⎜ , ⎟ .⎝ 2 2 ⎠9. Be able to find the distance from a point to a line. The distance from a point (x,y)to a line ax + by + c = 0 isax1+ by1+ cd = .2 2a + b

10. Know how to find the equation of a circle in applied situations.a. To find the center, find the midpoint of the diameter.b. To find the radius, find the distance from the center to a point on the circleor the distance from the center to a tangent line.c. After finding the center (h, k) and the radius r, substitute those numbersinto the standard equation of a circle: (x – h) 2 + (y – k) 2 = r 2 .11. Be able to find the properties of an ellipse.a. The two fixed points are called foci (singular is focus).b. The axis that contains the two given points called foci is the major axisand has a length of 2a. The other axis is the minor axis and has a length of2b.c. The points where the ellipse cuts its major axis are called its vertices(singular is vertex).d. The distance from the center to a focus equals c.e. The chord through either focus perpendicular to the major axis is called2b 2the latus rectum (LR) and its length is equal to .af. The eccentricity (e) is a number that indicates the degree of circularity;e = ac .g. The domain is the x-values of the major axis (position I) or minor axis(position II).h. The range is the y-values of the minor axis (position I) or major axis(position II).Ellipse in position I:12. Know how to write the standard equation of an ellipse (a is half of the length ofthe major axis and b is half of the length of the minor axis).a. If the foci are on the x-axis, the ellipse is in position I. The standardequation of an ellipse in position I with its center at the origin2 2x yis + = 1.2 b2a

. If the foci are on the y-axis, the ellipse is in position II. The standardequation of an ellipse in position II with its center at the origin2 2x yis + = 1.2 a2bc. The standard equation of an ellipse in position I with its center at (h, k)is ( ) 2( ) 2x − h y − k+ = 122a bd. The standard equation of an ellipse in position II with its center at (h, k)is ( ) 2( ) 2x − h y − k+ = 122b a13. Be able to write the general equation of an ellipse. The general equation of anellipse is Ax 2 + Cy 2 + Dx + Ey + F = 0.14. Know how to change from the standard equation of an ellipse to the general form:a. Multiply the binomial squares to get rid of the parentheses.b. Simplify.15. Know how to change the general equation of an ellipse to the standard form.a. Group like terms.b. Complete the square.c. Simplify.16. Be able to find the equation of an ellipse in applied situations.17. Know how to find the properties of a parabola with its vertex at the origin (0, 0).a. The directrix is the fixed line of a parabola. The directrix intercept is(-p, 0). The directrix is x = -p (position I) or y = -p (position II).b. The focus is the fixed point of a parabola. The focus is (p,0).c. The vertex is the point at which the parabola changes direction. Thevertex is (0, 0).d. A parabola in position I will open either to the right or to the left. Thestandard equation of a parabola in position I is y 2 = 4px (opens to theright) or y 2 = -4px (opens to the left).e. A parabola in position II will open either upward or downward. Thestandard equation of a parabola in position II is x 2 = 4py (opens upward)or x 2 = -4py (opens downward).f. The latus rectum is the chord through the focus perpendicular to the axisof the parabola. The length of the latus rectum is |4p|.g. The domain is all of the possible x-values. If the parabola is in position I,the domain is x≥0 (opens upward) or x≤0 (opens downward). If theparabola is in position II, the domain is all real numbers.h. The range is all of the possible y-values. If the parabola is in position I,the range is all real numbers. If the parabola is in position II, the range isy≥0 (opens to the right) or y≤0 (opens to the left).

Parabola in position I with vertex at the origin:18. Be able to find the properties of a parabola with its vertex not at the origin (0,0).a. The directrix is the fixed line of a parabola. The directrix is the line thatis the same distance from the parabola as the focus.b. The focus is the fixed point of a parabola. The focus is p units from theparabola.c. The vertex is the point at which the parabola changes direction. Thevertex is (h, k).d. A parabola in position I will open either to the right or to the left. Thestandard equation of a parabola in position I is (y-k) 2 = 4p(x-h) (opens tothe right) or (y-k) 2 = -4p(x-h) (opens to the left).e. A parabola in position II will open either upward or downward. Thestandard equation of a parabola in position II is (x-h) 2 = 4p(y-k) (opensupward) or x 2 = -4py (opens downward).f. The latus rectum is the chord through the focus perpendicular to the axisof the parabola. The length of the latus rectum is |4p|.g. The domain is all of the possible x-values. If the parabola is in position I,the domain is x≥h (opens upward) or x≤h (opens downward). If theparabola is in position II, the domain is all real numbers.h. The range is all of the possible y-values. If the parabola is in position I,the range is all real numbers. If the parabola is in position II, the range isy≥k (opens to the right) or y≤k (opens to the left).Parabola with its vertex at (h, k):19. Be able to graph a parabola.a. Find the vertex (h, k).

. Use the equation to find other points on the graph.20. Know how to write the equation of a parabola in standard form and general form.a. The standard equations of a parabola are(x-h) 2 = 4p(y-k) and (y-k) 2 = 4p(x-h)b. The general equation for a parabola isAx 2 + Dx + Ey + F = 0 or Cy 2 + Dx + Ey + F = 0.c. To change from standard form to general form, multiply the binomials toget rid of the parentheses, then simplify.d. To change from general form to standard form, complete the square, thensimplify.21. Be able to use the properties of the parabola to solve application problems. Thisincludes finding the locus of points. The locus of points is the set of all pointsthat satisfy a given condition.22. Know how to find the properties of a hyperbola:a. Asymptotes are the diagonal lines of the hyperbola that approach thehyperbola’s curve but never touch it.b. The conjugate axis is the line segment of length 2b of a hyperbola.c. The transverse axis is the line segment of length 2a of the axis that cutsthe hyperbola.d. The standard form of a hyperbola in position I is22( x − h) ( y − k)− = 122a be. The standard form of a hyperbola in position II is22( y − k)( x − h)− = 122a bf. The vertices of the hyperbola are at ± a.g. The center of the hyperbola is at (h, k).h. The eccentricity is ac and is greater than 1.2b 2i. The latus rectum is a.j. The equations of the asymptotes are y = ± ab x.

Hyperbola in position I:23. Be able to translate points in the Cartesian coordinate plane. Translation is arectangular movement of the graph.a. Use the formula x' = x – h to find the new x-value, where x is the originalx-value and h is the x-value of the new center.b. Use the formula y' = y – k to find the new y-value, where y is the originaly-value and k is the y-value of the new center.c. The new point is (x', y').24. Know how to find the center of a conic section and translate the center to theorigin.a. Complete the square and use the standard equation of the conic section tofind the center (h, k).b. Let x' = x – h and y' = y – k and replace x - h with x' and y - k with y' inthe equation of the conic section.c. Simplify.25. Be able to find the image of a point with respect to a rotation.a. Use the formula x ' = x cosθ + y sinθ to find the new x-value, where x isthe original x-value, y is the original y-value and θ is the measure ofrotation.b. Use the formula y ' = x sinθ - y cosθ to find the new y-value, where x is theoriginal x-value, y is the original y-value and θ is the measure of rotation.c. The image point is (x', y').26. Know the general form of the conic equation:Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0.27. Be able to identify the type of conic section from the equation.a. A circle has the same coefficient in front of x 2 and y 2 and no xy term.

. An ellipse has different coefficients with the same sign in front of x 2 andy 2 and no xy term.c. A parabola has only one x 2 or y 2 term and no xy term.d. A hyperbola has different coefficients with opposite signs in front of x 2and y 2 .