Find an equation for the ellipse. center at (5,6), focus at (11,6); contains the point (5,10)?
958287
(x - h)2 / a2  + (y - k)2 / b2 = 1  is the equation for an ellipse at center (h,k).  So (h,k) =(5,6)\n\nThen just plug in the other values you know and solve.  Since (5,10) is on the ellipse, x=5, y=10.\n\nSo we have (0-0)2/a2  + (10-6)2 / b2 = 1\nThen b2 = 16\n\nThe fact that the focus is at (11,6) is the clue to give you the a2 value.  \nThe center at (5,6) indicates that the other focus is (-1, 6)\nUse the pythagorean theorum to find the distance from each focus to (5,10) and add them together.\n\nSquare root (( 5+1)2 + (10-6)2) + Square root ((11-5)2 + (6-10)2) = Sqr( 36 + 16) + Sqr(36 + 16) \n= 2 Sqr (52)\n\nso there are two points on the ellipse on the horizontal axis of the ellipse, say one is (n,6).  \nThe distance from (n,6) to (-1,6) plus the distance from (n,6) to (11,6) = 2 Sqr (52)\n\nSqr ((n+1)2 + (6-6)2)) + Sqr ((n-11)2 + (6-6)2 = 2 Sqr (52)\nSqr (n+1)2 + Sqr (n-11)2 = 2 Sqr (52)\nn+1 + n-11 = 2 Sqr (52)\n2n -10 = 2 Sqr (52)\nn = Sqr(52) +5\n\n"a" is the distance from the center (5,6) to (n,6)...the length of the horizontal axis of the ellipse\n\n5 - n = Sqr (52) = a\n\nSo the equation for the ellipse is:\n\n((x-5)2/ 52) + ((y-6)2/16) = 1
1099920
I'd bet you can find it on this site:\n\nhttp://mathworld.wolfram.com/Ellipse.html
587217
First, always make the side with the squared term a perfect square so the equation is in standard form.  This is done by moving the constant term (25) to the left side so you have x - 25 = y^2 - 14y.  Now, complete the square on the left and your equation now becomes\nx - 25 + 49 = y^2 - 14y + 49 => x + 24 = (y - 7)^2.  It is of the form (x - h) = 4p(y - k)^2.  Whenever a parabola is written in this form, the vertex is always (h,k), or in this case, (-24,7).  For your equation, 4p = 1, so p = 1/4.\nSince the parabola opens to the right, the focus is 1/4 unit to the right of the vertex, so its coordinates would be (-24 + 1/4,7) = (-23 3/4,7).  The axis of symmetry is found by setting the squared term in your equation to 0 and solving for the variable.  In this case, 4p(y-7)^2 = 0 => y = 7 is the axis of symmetry.  The latus rectum is always 4p units long and in this case, 4(/14) = 1.  The latus rectum is a chord that goes through the focus and is perpendicular to the axis of symmetry.  Finding the direction of opening is simple.  Write the equation in standard form as I did above and make sure linear term is positive, then observe the following.  If the equation looks like (x - h) = 4p(y - k)^2 and 4p is positive it opens to the right, if 4p is negative, it opens to the left.  If the equation looks like (y - k) = 4p(x - h)^2 and 4p is positive it opens up, if 4p is negative, it opens down. In all cases, the focus is always p units from the vertex and the latus rectum is always 4p units long and the vertex is always (h,k).  Let me know if this is not clear.
70526
=((-4y)^2)/(10y-5)= (16y^2)/(10y-5)\nand if the equation is equal to zero then y=4i
960038
Since the foci of this hyperbola both lie on the horizontal line y=7, then the major axis is horizontal and the minor axis is vertical, and so the equation for it is \n\n((x-a)^2)/(M^2) - ((y-b)^2)/(m^2) = 1 \n\nwhere the center of the hyperbola (the midpoint of the segment between the two foci) is (a,b), M is the length of the semi-major axis (half the distance between the vertices),  and m is the length of the semi-minor axis.  If d is half the distance between the foci, then M^2 + m^2 = d^2\nThe center of this hyperbola is (7,2) since it is the midpoint of the segment between the foci.  M = 3 (the distance from the center to each vertex), d = 5 (the distance from the center to each focus), and so m = 4 since 3^2 + 4^2 = 5^2.  \n\nThus this hyperbola has equation \n\n((x-7)^2)/9 - ((y-2)^2)/16 = 1
1078325
Let as suppose that your ellipse is rotated by 45 degrees to right (or by -45 degrees). The new coordinates are X = x cos (-45) - y sin (-45) and Y = x sin (-45) + y cos (-45) and they are satisfying X^2/a^2 + Y^2/b^2=1. There for (x + y)^2/2a^2 + (x - y)^2/2b^2 = 1 (in a few easy steps). The mayor axis over the x=y is given by |a| > |b| .
1180944
alright, ill tell you, but not without explaining it.  im giving you the benefit of the doubt that you are not just cheating on your homework :)\n\nthe standard equation of a circle, using only variables is\n(x-h)^2 + (y-k)^2 = r^2\nwhere the center is at (h,k) and the radius is r\n\nthat means that when you look at the equation youre given, the x coordinate of the center is whatever number is subtracted from the x\n\n(x+1) ---> (x- -1)  so  h = -1\n\nyou use the same process to determine "k", which is the y coordinate of the center\n\n(y+3) ---> (y- -3)  so  k = -3\n\nthat makes the center of the circle at (-1, -3)\n\n\nfinding the radius is also simple using the standard equation. here it is again, just for reference:\n(x-h)^2 + (y-k)^2 = r^2\n\nsee how you can use that to get this?\n9/4 = r^2\n\nso just take the square root of both sides . . .\nr = +/- 3/2\n\nbecause you took the square root, the 3/2 could technically be either positive or negative, but since youre talking about the radius of a circle and you can't have a negative distance, you only take the positive answer.\n\nso r=3/2
387856
All these problems are available in coordinate geometry text books.  For the problem 1)\n\nThe equation of the circle with A(x1,y1)and B(x2,y2) as end points of diameter is (x-x1)(x-x2) + (y-y1)(y-y2) = 0\n\nSimply substitute A(2, -3) and B(8,5) wou get \n(x-2)(x-8) + (y+3)(y-5) = 0\nthat is, x^2 - 10x+16 + y^2 -2y -15 =0\nOr, x^2 + y^2 - 10x -2y + 1 =0\n\nThis is the required equation.\n\n2) In standard form of parabola y^2 = 4ax, the origin (0,0) is vertex, S(a,0) is focus and x + a = 0 is the equation of directrix.  In the present case, the focus is at (2,3).  this means the coordinates of vertex are (0,3)and the equation of diretrix is x + 2 = 0.  Therefore the equation of the parabola is (y-3)^2 = 4. 2. (x-2)\nOr y^2 -6y -8x + 25 =  0\n\nSimilarly attempt the problems 3 & 4
183039
The formula for a circle is (x-h)^2 + (y-k)^2 = r^2, so your center is (1,4).  The radius squared is 9, so the radius is 3.  So from (1,4) go up 3, down 3, left 3 and right 3.  Then connect the dots in a circle.
1009010
y= 40-5x\ny=40-5(6)\ny=10\n10
1010902
This is  basic\nslope= m=(y2-y1)/(x2-x1)=(12-7)/(8-6)=5/2\nangle=arctan(5/2)
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
461911
The equation is (x-h)^2 + (y-k)^2 = r^2, \nwhere (h,k) is the center and r is the radius.\nTHe center is (0,0), so that means h = 0 and k = 0.\nThe radius is 10, so that means r = 10\nPlug these values into the equation.\n(x-0)^2 + (y-0)^2 = 10^2, which simplifies to\nx^2 + y^2 = 100
869748
It's a circle of radius 4, centered at the origin.\n\nThe equation of a circle centered at (a, b) is:\n\n(x - a)² + (y - b)² = R²\n\nWhen the circle is centered at the origin, (a, b) = (0, 0), or:\n\nx² + y² = R²\n\nTherefore, y is simply given as:\n\ny = ±sqrt( R² - x² )\n\nIn this case, R² = 16, so R = 4. The plot you want to make is one such that every point on the plot is 4 units away from the origin.
524037
The general form of the equation of a circle is\n\n(x - a)^2 + (y - b)^2 = r^2\n\nwhere (a, b) is the center and r is the radius.  So for your circle,\n\n(x - 4)^2 + (y - 7)^2 = 36\n\nthe radius is 6.  (And FYI, the center is at (4, 7).)\n\nEDIT:  For your second question, the answer is false - the coordinates would be (3, -5).
822973
Since the foci are along the y-axis, the major axis of the ellipse will be along the y-axis.  Also, the center of the two foci is the center of the ellipse, which will be (0,0) or the origin.\n\nSo the only thing left to figure out is the length of the minor axis.\n\nThe major axis, minor axis, and focal length (distance from the foci to the center) are related by the Pythagorean theorem:\n\na² + b² = c²\n\nYou know that a = 26/2 = 13, c = 12, so b = 5.\n\nTherefore the equation is:\n\nx²/25 + y²/169 = 1
207955
u need more info than just the length, but here is something which might help u are speaking of ovals( called ellipses in math)\n\n\nwell the formula for them is\n\n\n(x^2)/a^2  +  (y^2)/b^2  =1\n\nwhere a and b are half of the major and minor axis , you only know 1 of these meassures, thats not enough sorry chap, wish i could help, but i think no one can, with only that info