Math for the 2nd term
Go to Long Test II

Part 1: Quadratic Equations in one variable	Quadratic Equations deal with polynomials written with variables in the 2nd degree. Because the variables are in the 2nd degree, we expect two answers (or roots) for every equation. The standard form for quadratic equations is:       Ax2 + Bx + C = 0 In this arrangement A ≠ 0 because if A = 0 then the polynomial is not in the 2nd degree and we would have only a linear equation. Balik nnman first term. The ways of solving quadratic equations are listed below:
Zero-factor property	Basically solve by factoring. The key assumption here is that if ab = 0, then either a or b = 0. Oonga naman, kung multiply mo sa 0 kaht anong no. ay 0 ang sagot mo so either a or b is zero. That's why we can equate each factor of the quadratic expression to 0. So if we factor the expression          Ax + By + C = 0 makukuha natin ung (ox+p)(qx+r) where o, p q and r are real nos. Next step is to equate both to zero. ox + p = 0 and qx + r = 0 Our two answers are: x = -p/o and x = -r/q Nakuha ang sagot na yan by transposing the p and the r. Then, to remove the coefficient of x we divide by o or q respectively. Wag kalimutan isulat ang solution set.
Square-root property i	There are two square-root property possibilities. Eto ung una. This is a quick way of finding the value of x, if and only if the quadratic equation is in the form          x2 = k The solution is x = ±√k may ± dahil pag ni-square ang negative no. lalabas ay positive parin. Note that if k is a negative number, there is no real solution to the equation.
Square-root property ii	Second square root property. The quadratic equation is written in the form          m(px + q)2 = r Ang solution dito first divide r/m to get rid of m on the left, then get the square root. √(px + q)2 = ±√r/m the left becomes simply px + q and the right, resolve it. tapos use algebra to isolate x, to get the values.
Completing the square	This method also uses factoring as the main solution.   First write the quadratic equation in the form (x2 + Bx + C = 0).   Then take the value of B and divide by 2. Then square the quotient.   The value of the squared quotient is the value that your C is supposed to be in order to complete the square. Add it to both sides of the equation   Once you have the perfect square trinomial you can factor it out. Use the factoring tools such as difference of 2 squares, sum of 2 squares, etc.
Quadratic Formula	This method is longer than the other special methods, but it is a sure way to find the value of the variable in any quadratic equation. It will work with all quadratic equations.     First write the equation in standard form. Then use the formula: x = -b ± √bČ - 4ac        2a

Part 2: More on Quadratic Equations	Dito isiningit ang special lessons.. more on the discriminant value here, how it relates to the roots of an equation, and the sum and product of the roots.
Discriminant value and type of roots	Depending on the discriminant value that you get, you will be able to determine the kind of roots the quadratic equation will have. Refer to the table. the discriminant is equal to b2 -4ac Discriminant value Types of roots 0 double root (mult. 2) perfect square 2 real rational (pde fractions) positive but not perfect square 2 real irrational (may √) negative 2 imaginary nos.
Sum and product of roots	The sum of the two roots of an equation is given by the expression negative b over a. In this case, a is the coefficient of the square of x, tapos b is the coefficient of the non-squared x. The product of the roots is given by the expression c over a. Use a as above, while c is the constant at the end of the quadratic equation (in standard form). Mahahanap ang roots, or the equations themselves using these two expressions. If given the exact root values, you can add them and multiply them, tapos equate to the c/a or -b/a to get the standard form of the quadratic equation. If given the standard form, you can divide by a, get the negative of b/a to get the sum of the roots. Same with the product, use c/a.

Part 3: Conic Sections	*Intersection of a plane with two right vertical cones. Given by a quadratic equation in two variables. General form is in: Ax2 + By2 + Cy     Intersections take on 4 shapes.
Introduction	Laking pasalamat ko sa first Math project for the term kase hindi ko na kailangan idrawing ung conic sections (sa paint! argh!).. more or less napi-picture na ang itsura ng conics dahil dun. So conic sections show intersections. since the plane extends infinitely, nagkakaroon minsan ng plane that intersects the cones twice. Yun ung hyperbola. as for the other three, kahit na gaanong kalayo mo iextend ung planes, hanggang dun lang sila magi-intersect with the plane. the three other shapes are the ellipse, circle and parabola. The altitude runs in the center of the cones, it is an imaginary vertical line.
Ellipse	Diagonal intersection of a plane with the cones.
Circle	Special occurence of an ellipse. Occurs when the plane intersects with the cones horizontally. Plane is perpendicular with the altitude of the cones
Parabola	Song ng Tool. De joke lang, pero may song tlga ang tool na parabola. Sa math naman, this is the Intersection of the plane with the edge of the cones. Makes a sort of curved shape
Hyperbola	Vertical intersection of a plane with the cones. Parallel with the altitude of the cones.
How to tell w/c conic from the gen. form	This next table shows the conditions of a certain conic given its general form. Para malaman mo na agad what kind of conic pala tlga un. It was given nung lt2, medyo nakakagulo din. General form of all conics: Ax2 + By2 + Cx + Dy + E = 0 Conic Condition Example Parabola A = 0 or B = 0 (only 1 quadratic) 4y2 + 2x + y + 7 = 0 Circle AB > 0 and A = B (A and B same sign, equal) 6x2 + 6y2 + 4x + 24y + 100 = 0 Ellipse AB > 0 and A ≠ B (A and B same sign, unequal) 25x2 + 9y2 - 100x - 90y + 100 = 0 Hyperbola AB < 0 (A and B have opposite signs) 25x2 - 9y2 - 100x - 90y + 100 = 0

Part 4: PARABOLA

The parabola. ung unang conic na tinalakay in detail. it's a curve. many practical applications in real life.

Definitions

Set of all points equidistant from a fixed point (called the focus) and a line below its vertex called the directrix. There are 4 possible orientations of a parabola: up, down, left and right Quantities Vertex: The point along the axis of symmetry where the arms of the parabola originate Axis of symmetry: Line passing through the vertex separating the parabola into symmetrical parts Focus: Fixed point along the focal chord and the axis of symmetry Focal chord: or latus rectum, line segment parallel to directrix which contains the focus and is cut off by the parabola Directrix: line behind the vertex. Aside from the focus, the points of the parabola are equidistant to this line. p = length from the vertex to the focus. This is also equal to the length from the vertex to the point on the directrix along axis of symmetry 4p = the length of the focal chord Standard Form Equation (x - h)2 = ±4p(y - k) (y - k)2 = ±4p(x - h) Orientation Up/Down Left/Right Vertex (h, k) Positive p (concavity) Up Right Focus (h, k + p) (h + p, k) Directrix y = k - p x = h - p General Form Ax2 + Bx + Cy + D = 0 Ay2 + By + Cx + D = 0 Graph Negative p (concavity) Down Left Focus (h, k - p) (h - p, k) Directrix y = k + p x = h + p General Form Ax2 + Bx + Cy + D = 0 Ay2 + By + Cx + D = 0 Graph

Start of Long Test II

Part 5: CIRCLE	"Circle" - song ng Slipknot sa 3rd album.. acoustic song un. ("all that i wanted, were things i had b4...")
Definitions	Set of all points equidistant from a fixed point called the center of the circle. the fixed dist. is called the radius of the circle. Twice the length of the radius is called the diameter of the circle. This is a line segment connecting two points on the circle and passing thru the center. Standard form is in equation:    (x - h)2 + (y - k)2 = r2   where    C(h,k) and r = radius General form is in equation: Ax2 + Ay2 + Bx + Cy - D = 0   (coefficients of quadratic variables are equal)
Solving Given Standard form	(x - 1)2 + (y + 4)2 = 36 You're given the standard form, solve the circle Center (1, -4) From (h,k) Radius 6 from r2 = 36; just get the square root Graph Center Circle Radius Get General Form (x - 1)2 + (y + 4)2 = 36 Start with standard. write (a2 + 2ab + b2) form x2 - 2x + 1 + y2 + 8y + 16 = 36 Transpose 36 and combine with 1 and 16 x2 - 2x + y2 + 8y - 19 = 0 Final.
Solving given General Form	x2+4x+y2+4y-12 = 0 Start by getting the standard form. Get standard form x2+4x+y2+4y=12 complete da skweyers (squares) (x2+4x+4)+(y2+4y+4)=12+8 Add 4 to x, 4 to y and 8 to other side. factor (x + 2)2 + (y + 2)2 = 20 Final Center (-2, -2) Quadrant III yan.. both negative Radius √20 from r2 = 20; approx. 4.4 something Graph Center Circle Radius
Solving given Endpoints of a Diameter	Actually really easy. Use the midpoint formula to get the center and use the distance formula to get the radius. Since we know that the diameter passes thru center, automatic midpt ng endpts ay center. (3, 6) and (1, 0) are endpoints of a diameter Use midpoint formula to get center x1 + x2 ,     y1 + y2     2                  2 This is the basic midpoint formula. Parang average ng x tyka average ng y. parang. 3 + 1 ,    6 + 0     2           2 The resulting x and y coordinates ang coordinates ng center (2, 3) Eto na ang center! (h, k) Use dist. formula to get the radius r2 = (x - h)2 + (y - k)2 Distance formula is from a point to the center, since radius.. not from point to point, kac dat would giv u the diameter length, not d radius. r = √(3 - 2)2 + (6 - 3)2 Use any of the 2 endpts, and the radius. Dito ginamit ang (3, 6) na point. Get the square root of this para r ang matira sa other side. r = √10 yan na ung radius length. ayus! Standard form (x - 2)2 + (y - 3)2 = 10 From (x - h)2 + (y - k)2 = r2 General form x2-4x+4+y2-6y+9-10=0 Combine like terms x2+y2-4x-6y+3=0 Final Graph Center Circle Endpts of diameter

Part 6: ELLIPSE	Alam nyo ba ung elliptical road? ung papuntang UP-D??? Eh track oval, nakakita na kyo?
Definitions	Set of all points wherein the sum of the distances from 2 fixed points (called foci) is constant. Has 2 vertices, a center, a major axis and a minor axis. Come to think of it, ang ellipse and hyperbola pala sobrang alike. Ung hyperbola nga lang parang ellipse na biniyak sa gitna tapos hinarap outward. Mas nauna ko kac ginawa ung hyperbola portion kac may quiz na bout that one. Here's a picture to help visualize the definition and concept of an ellipse. Makikita ditow na merong 2 fixed points na wla sa surface ng ellipse. These are the foci. Tas ung dot na gumagalaw represents any point on the ellipse. The red lines are distance of that pt to focus 1, tyka distance niya from focus 2, kaya dalawa ung lines. For each pt nagiiba ung value ng dalawang distance na to. Pero when you add them, the sum of the distances ay pareho lang for all pts. Pag lumaki ung dist 1, siguradong liliit ung dist 2 and vice versa. That keeps the sum dist constant. pic from mathworld.wolfram.com Standard form of equation (x - h)2 + (y - k)2 = 1    a2             b2 (x - h)2 + (y - k)2 = 1    b2             a2 Center (h, k) Vertices (h ± a, k) (h, k ± a) Major Axis Horizontal Vertical Minor Axis Vertical Horizontal Foci (h ± c, k) (h, k ± c) Focal length (c) (f1 to f2) c2 = a2 - b2 Eccentricity of ellipse  c  a -a measure of "roundness" Illustration General form of equation Ax2 + By2 + Cx + Dy + E = 0 where A and B have the same sign Other Conditions/values major axis: line segment from V1 to V2 minor axis: line segment | to major axis passing through center and cut off by ellipse a = length of semi-major axis (1/2 length of entire major axis) [blue line] b = length of semi-minor axis (1/2 length of entire minor axis) [red line] c = length from center to either focus (1/2 of focal chord) [green line] a > b    (a is always greater than b). This condition helps to identify which quantity ang denominator ng what coordinate (x or y) sa standard form. Para alam mo na agad ang shape of ellipse, look for the larger no. eccentricity doesn't really come out a lot sa tests. basta't c/a sya dat's all u hav 2 remember ok na un.
Solving given standard form	(x - 2)2 + (y + 1)2 = 1     36            4 Find the elements involved. 36 is larger than 4 so automatic ang ellipse ay flat/ horizontal dahil a is associated with x. Center (2, -1) Na-establish natin na a is larger than b. so if a is under x then the ellipse is a horizontal one, kac mas mahaba ung x length nya kesa sa y length. a2 = 36 Then a = 6 b2 = 4 Then b = 2 Vertices (8, -1) Since horizontal, use (h ± a, k) to find vertices (-4, -1) c2 = 36 - 4 Then c = √32 This is between 5 and 6, approaching 6 Foci (2 - √32, -1) No need to resolve/compute the exact value (2 + √32, -1) Graph Center Vertices Foci Ellipse a (semi-major axis) b (semi-minor axis) c (half focal chord) Get General Form (x - 2)2 + (y + 1)2 = 1     36            4 Start with standard form. Get multiple to cancel 36 and 4 (36). Multiply both side (x - 2)2 + 9(y + 1)2 = 36 write in (a2 + 2ab + b2) forms. distribute 9 to y x2 - 4x + 4 + 9y2 + 18y + 9 = 36 transpose 36 and combine like terms x2 + 9y2 - 4x + 18y - 23 = 0 Final
Solving given general form	25x2+9y2-100x-90y+100=0 Start by solving for the standard form. Group into x and y Get Standard Form 25x2-100x+100+9y2-90y=0 100 was grouped with x because it's factorable. 25(x2-4x+4)+9(y2-10y+25)=225 naging 225 kasi mult. by nine sa labas. factor 25(x - 2)2 + 9(y - 5)2 = 225 Divide both side by 225 (x - 2)2 + (y - 5)2 = 1     9             25 Final. Get components and graph. Automatic vertical ung ellipse kac 25 is d bigger no. and is associated with y. Center (2, 5) ayoko maging repetitive at redundant. klaro naman dba? ;) a2 = 25 Then a = 5 b2 = 9 Then b = 3 Vertices (2, 10) Since vertical, use (h, k ± a) to find vertices (2, 0) c2 = 25 - 9 Then c = 4 From the square-root of 16, difference ng 25 at 9 Foci (2, 9) From 5 ± 4 (2, 1) Graph Center Vertices Foci Ellipse a (semi-major axis) b (semi-minor axis) c (half focal chord)

Part 7: HYPERBOLA	Para siyang ellipse.. un nga lang opens outward, not inward
Definitions	Set of all points where the difference of the distances from 2 fixed points (called foci) is constant. Has two vertices constituting the endpoints of a transverse axis which passes thru its center. Standard form given by the equations: Standard form equation (x - h)2 _ (y - k)2 = 1   a2              b2 (y - k)2 _ (x - h)2 = 1   a2              b2 Center (h, k) Vertices (h ± a, k) (h, k ± a) Transverse axis and Hyperbola Horizontal Vertical Conjugate axis Vertical Horizontal Foci (h ± c, k) (h, k ± c) Focal length (c) c2 = a2 + b2 Equation of Asymptotes y - k = ± b/a (x - h) y - k = ± a/b (x - h) Auxillary Rectangle (h ± a, k ± b) (h ± b, k ± a) Illustration General form Equation Ax2 + By2 + Cx + Dy + E = 0 where A and B have opposite signs Legend a = length from transverse axis to a vertex (blue line) b = half the length of conjugate axis (axis perpendicular to transverse axis passing thru center) (red line) asymptotes - lines formed by the diagonals of the auxillary rectangle and determine the curve or eccentricity of the hyperbola (green lines) auxiliary rectangle - formed by extreme corners of b and a (purple)
Things to remember	Sa standard form, a palagi nasa left. Meaning the positive var. is always associated with a, at ang negative variable lagi nasa b. Make sure you know which coordinate (x or y) is associated with which axis (a or b). whatever's associated with a will tell the direction of the opening. x is left/right and y is up/down. Don't bother using the equations of asymptotes to find the slope of the line, y-intercept, etc. just connect the corners of the rectangle with diagonal lines and extend In graphing, remember that the curves of the hyperbola must not touch the asymptotes, pero it approaches the asymptotes very nearly Make sure you know when you're dealing with a, 2a or a2. Also with b, 2b or b2 and specially when to use a/b as opposed to b/a To get standard form from general form complete the square (sabay, one for x, one for y) Kung may coefficient sa labas ng parentheses while completing the square for example 5(x2 + 6x), it becomes 5(x2 + 6x + 9). What you add on the other side ng = sign ay 9·5 kase u added 9 times the 5 coefficient sa labas ng parentheses bale 45.
Solving given standard form	(y - 2)2 _ (x + 1)2 = 1    9               25 Find the elements of the parabola by solving. a is associated with y so automatic alam mo na na up/down ung hyperbola. Center (-1, 2) As u can see, the hyperbola is a vertical one and b is larger than a. Ok lang to. Unlike the ellipse, wlang major o kaya minor axis, no rule that says a should be greater a2 = 9 Then a = 3 b2 = 25 Then b = 5 Vertices (-1, -1) Since vertical, you use the (h, k ± a) to find the vertices so its (-1, 2 ± 3). (-1, 5) c2 = 9 + 25 Then c = √34 Square root of 34 is near 6 but less than 6. Foci (-1, 2 - √34) Since we don't have exact values, accepted na yang sagot na yan.. may √, rem na ang foci (h, k ± c) (-1, 2 + √34) Asymptotes y - 2 = 3/5 (x + 1) Since up/down ung hyperbola, use a/b NOT b/a. Ang fraction na yan ang slope ng asymptotes and since ung transverse axis vertical, a ung rise niya at b ung run nya y - 2 = - 3/5 (x + 1) Graph Center Hyperbola Asymptotes Auxillary Rectangle b a Get General Form (y - 2)2 _ (x + 1)2 = 1    9               25 Start with standard form. Get multiple to cancel both 9 and 25 (9 * 25 = 225). Multiply to both sides 25(y - 2)2 - 9(x + 1)2 = 225 Then write the squares of binomials into (a2 + 2ab + b2) form 25(y2-4y+4)-9(x2+2x+1)=225 Distribute 25 to y and 9 to x 25y2-100y+100-9x2+18x+9=225 Transpose 225 and combine like terms -9x2+25y2+18x-100y-116=0 Final
Solving given General Form	x2-4y2+4x-24y-48=0 Solve for all components of hyperbola Get Standard Form x2+4x-4y2-24y-48=0 Transpose 48, group x's together, y's together and try to complete the square (x2+4x+__)-4(y2+6y+__)=48 4 was factored out from the y side first. Complete the square (x2+4x+4)-4(y2+6y+9)=48+4-36 add what's necessary on both sides. -36 was added kac 9 times the -4 sa labas.. un ung sa things 2 remember no. 6. factor the trinomials (x+2)2 - 4(y+3)2 = 16 Divide both side by 16, that should give you the standard form (x+2)2 _ (y+3)2 = 1   16           4 Final. You can now start solving for the components Center (-2, -3) Hyperbola is horizontal this time and a is larger than b. a2 = 16 Then a = 4 b2 = 4 Then b = 2 Vertices (-6, -3) Use (h ± a, k) for vertices this time (2, -3) c2 = 16 + 4 Then c = √20 Square root of 20 is between 4 and 5 Foci (-2 - √20, -3) Foci have coordinates (h ± c, k) for a horizontal hyperbola (-2 + √20, -3) Asymptotes y +3 = 1/2 (x + 2) Now we use b/a kac naging a naman ung run at b ung rise. ang makukuha mo actually ay 2/4 pero lowest term nun ay 1/2. y + 3 = - 1/2 (x + 2) Graph Center Hyperbola Asymptotes Auxillary Rectangle b a

© 2004 by Luis Medina. Unauthorized duplication and distribution of this document will be persecuted and is punishable by federal law.... ayus!