# Questions on Algebra: Conic sections - ellipse, parabola, hyperbola answered by real tutors!

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 Algebra: Conic sections - ellipse, parabola, hyperbola Solvers Lessons Answers archive Quiz In Depth

 Question 482101: A standard form equation for the ellipse with foci (0,+ or - 3)and major axis length 10 is? Click here to see answer by lwsshak3(6513)

 Question 482730: Find the equation of a hyperbola whose asymptote coincide withe the points (-7, 4) and (-1, -14). Its other asymptote contains point (-4,1), and (-3/4, -53/4) is a point on the hyperbola. Click here to see answer by lwsshak3(6513)

 Question 483182: how do i graph this and make a parabola? g(x)=x2-9x+9 Click here to see answer by edjones(7569)

 Question 482543: Explain which conic section this equation and explain how to solve it: 12x^2-18y^2-18x-12y+12=0 Click here to see answer by lwsshak3(6513)

 Question 484442: Hello, I need a quick brush-up on ellipses: find center of ellipse x^2+4y+6x-8y+9=0 mainly i just need to know how to get this in the right form, and then I think I'm good from there on figuring out the center, etc. I completed the square and got (x+3)^2+4(y-1)^2=1, but there is still the 4 in front of the (y-1)^2 - how do i get rid of the 4 while keeping a 1 on the right side? Thanks, Jessica Click here to see answer by lwsshak3(6513)

 Question 485884: Find the equation for the parabola with vertex (0, 0) and directrix x = 4 Click here to see answer by lwsshak3(6513)

 Question 485440: how the shape of the parabola relates to the equation. Click here to see answer by lwsshak3(6513)

 Question 486178: how do I find an equation of the parabola that contains the points (-3,7), (2,9), (6,-2) Click here to see answer by solver91311(16897)

 Question 486333: please help me finish this hyperbola: this is what it gives me. The hyperbola with center at (5, -6), with vertices at (0, -6) and (10, -6), with (15, 0) a point on the hyperbola. here is the equation setup: ( )^2/( )-( )^2/( )=1 Click here to see answer by lwsshak3(6513)

 Question 486326: (x-3)^2/9-(y-2)^2/16=1 I have tried everything and cant not figure out a couple of my college algebra questions. I am looking for the CENTER and VERTICES of this hyperbola. Please help me! Click here to see answer by lwsshak3(6513)

 Question 486510: Hello, I am having a problem finding the equation for a hyperbola. I have tried to work it out but can't seem to get it 100 percent right. The information my teacher gives is : The hyperbola with center at (4, 5), with vertices at (0, 5) and (8, 5), with b = 2. If you could please help show me how to solve this so i know how to do it that would be great. Thankyou so much. Click here to see answer by lwsshak3(6513)

 Question 472587: Find the equation in standard form of an ellipse with center at (0,0) minor axis of length 18, and foci at (0,-12) and (0,12). a. (x^2/225)+(y^2/81)=1 b. (x^2/144)+(y^2/81)=1 c. (x^2/81)+(y^2/225)=1 d. (x^2/9)+(y^2/15)=1 e. (x^2/15)+(y^2/9)=1 Click here to see answer by lwsshak3(6513)

 Question 482542: Explain which conic section this equation and explain how to solve it: 12x^2-18y^2-18x-12y+12=0 Click here to see answer by lwsshak3(6513)

 Question 487841: Find the vertex of -0.2x^2+12x+11 I have used two different methods and came up with two different answers. I have solved for the x= 30 but when I went to do the maximum value I used two solutions and that is where I came up with different answers. The first I substituted the 30 in to the equation -0.2(30)^2+12(30)+11= 191; on the second I used the formula 4ac-b^2 /4a with a= -0.2 b= 12 c=11 I got 4(-0.2)(11)-12^2 divided by 4(-0.2) and came to the answer 169. Click here to see answer by John10(245)
 Question 487841: Find the vertex of -0.2x^2+12x+11 I have used two different methods and came up with two different answers. I have solved for the x= 30 but when I went to do the maximum value I used two solutions and that is where I came up with different answers. The first I substituted the 30 in to the equation -0.2(30)^2+12(30)+11= 191; on the second I used the formula 4ac-b^2 /4a with a= -0.2 b= 12 c=11 I got 4(-0.2)(11)-12^2 divided by 4(-0.2) and came to the answer 169. Click here to see answer by nerdybill(6962)

 Question 471709: Find the equation in standard form of an ellipse with center at (0,0) minor axis of length 18, and foci at (0,-12) and (0,12). a. (x^2/225)+(y^2/81)=1 b. (x^2/144)+(y^2/81)=1 c. (x^2/81)+(y^2/225)=1 d. (x^2/9)+(y^2/15)=1 e. (x^2/15)+(y^2/9)=1 Click here to see answer by lwsshak3(6513)

 Question 491703: how do you find the equation of the parabola whose vertex is (4,-11) and that passes through the points (-6,16) Click here to see answer by stanbon(57361)

 Question 493252: Find the equation in standard form that has a vertex (-4, 7) and goes through (5, 0). Click here to see answer by lwsshak3(6513)

 Question 491838: graph the hyperbola and label the vertices, foci and asymptotes: (((y+2)^2)/25))-(((x+11)^2)/4)=1 Click here to see answer by lwsshak3(6513)

 Question 491835: find the center,vertices and foci of the ellipse: 16x^2+4y^2+64x-24y+36=0 Click here to see answer by lwsshak3(6513)

 Question 491832: find the ratius and center of the circle: x^2+y^2-8x+14y+29=0 Click here to see answer by lwsshak3(6513)

 Question 493986: Find the center and radius of the circle. 64x2 + 64y2 + 16x + 128y + 1 = 0 Click here to see answer by lwsshak3(6513)

 Question 495972: I know that the following equation will result in a circle graph (the teacher's key says so). But I would like to know how to get there. I need to isolate y: x^2 + y^2 - 4y = 21 (that is, x squared plus y squared minus 4y equals 21) Thank you Click here to see answer by nerdybill(6962)

 Question 496178: Give the solution of the following system. [y=x^2+4 y-4x=1] Click here to see answer by unlockmath(1602)

 Question 498056: Express f(x) in the form a(x - h)^2 + k. f(x)=x^2-6x+10 Click here to see answer by scott8148(6628)

 Question 498546: x2 − 2y = 8x −10 what is the focus and focus width of this parabola? Click here to see answer by lwsshak3(6513)

 Question 498267: writing an equation, in standard form, of the ellipse with the center at (-1,3), vertex at (3,3), and a minor axis of length 2. I have this so far (x+1)^2/( )+ (y-3)^2/( )=1 I don't know what to do with the minor axis??? Click here to see answer by lwsshak3(6513)

 Question 498944: what is the shape of the graph of the equation below? x^2/3+(y-4)^2/6^2=1 Click here to see answer by Theo(3464)

 Question 502968: Identify the graph of the conic section defined by the following equation. x^2+4x-y+9=0 Click here to see answer by lwsshak3(6513)

 Question 502801: if you are given a parabola and the focal diameter is 12, how do you find what the focus is? Click here to see answer by lwsshak3(6513)

 Question 502616: identify the center, vertices, length of the major axis and of the minor axis of each. then sketch the graph: 4x^2+y^2+40x-8y+112=0 Click here to see answer by lwsshak3(6513)

 Question 502428: Determine the coordinates of the foci of the hyperbola : y2-2x2-4x-4y=0. (Round off answer to nearest hundredths.) Click here to see answer by lwsshak3(6513)

 Question 503166: find the vertex, the line of symmetry and the maximum or minimum value of f(x). Graph the function f(x)=.2(x+6)^2+1 the vertex is= the line of symmetery is= what is the man/min of f(x)= is the value f(-6)=1, a min or max? then graph that represents. Click here to see answer by lwsshak3(6513)

 Question 500591: So this is where I am at with #s 1 and 2. Not sure if they were done correctly. The problem is that I have no idea how to graph a parabola or whatever. Please help me ANYONE!!I sure would appreciate it! Thanks.....SuzieFor questions 1-2, apply the quadratic formula to find the roots of the given function, and then graph the function. 1. (6 points) f(x) = x2 + 4 F(x) = x^2 + 4 Ax^2 + bx + c = 0 X =-b+√b^2-4*a*c 2*a X^2 + 4 = 0 Since there is no x term the 2nd coefficient is 0 X^2 + 0*x+4=0 A=1 b=0 c=4 X=0+√(0)^2-4*1*4 2*1 X =0+√0-4*1*4 2*1 X=0+√0+(-16) 2*1 X=0+√-16 2*1 X=0+4*i 2*1 X=0+4*1 2 X=0+2*I or x=0-2*i See Graph Attached We can see that there are no real roots 2. (6 points) g(x) = x2 + x + 12 G(x) = x^2 + x + 12 Ax^2 + bx + c = 0 X = -b+√b^2-4*a*c 2*a X^2 + x + 12 = 0 A=1 b = 1 c = 12 X = -1+√(1)^2-4*1*12 2*1 X = -1+√1-4*1*12 2*1 X = -1+√1+(-48) 2*1 X = -1 +√-47 2*1 X = -1+i*√47 2*1 X = -1+i*√47 2 X = -1-i*√47 2 2,18 -2,14 4,32 -4,24 Vertex is 0,12 Y = x^2 +x +12 SEE GRAPH ATTACHED we can see there are no real roots Click here to see answer by lwsshak3(6513)