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A standard Hyperbola has its vertices ar 5,7 and 3,-1.Find its equation and its asymptotes

What is the equation of a Standard Rect.Hyperbola given its vertices are at 5,7 and 3,-1/Also find its asymptots.

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hyperbola - Purplemath | Home


... the foci in an hyperbola are further from the hyperbola's center than are its vertices: The hyperbola ... hyperbola's equation, ... asymptotes ' equations ...
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Conics: Hyperbolas: Finding Information From the Equation


Explains and demonstrates how to find the center, foci, vertices, asymptotes, and eccentricity of an hyperbola from its equation. Also shows how to graph.
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equation of the hyperbola - Math Is Fun


... and for a hyperbola it is always greater than 1. The hyperbola is an open curve (has no ... two vertices (where each curve makes its ... The asymptotes are the ...
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jwilson.coe.uga.edu


1. Sketch the graph and find the vertices, ... An ellipse has its center at the origin. Find an equation of the ellipse ... (7, -1) and (-3, -1) (add/subtract 5 from ...
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Objective: Be able to find the equation, vertices, and foci of a


the focal axis and that has the center of the hyperbola at its ... Hyperbolas with Center (0,0) Standard Equation Focal axis x ... ( 1,2) and (5,2) and vertices 0 ,2 ...
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write the equation of the hyperbola in standard form: x^2-4x ...


... altitude line AR in standard form. A. x - y = 1 B. x + y ... asymptotes y= 18/5 x math find the equation of the ... equation for the hyperbola vertices (3 ...
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Precalc Check My Answer On Hyperbola And Vertices


Precalc Check My Answer On Hyperbola And Vertices . ... Find equation of hyperbola with asymptotes slopes ... standard Hyperbola has its vertices ar 5,7 and 3,1 ...
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A standard Hyperbola has its vertices ar 5,7 and 3,-1.Find its equation and its asymptotes

this equation expands toThe vertices share no common coordinate so the hyperbola cannot be orientated in the usual x-y form. The hyperbola must therefore be skew (tilted axis) and the line joining the vertices is sloping. This line is given by the equation y=((7+1)/(5-3))x+c where c is to be found. y=4x+c, and -1=12+c, plugging in (3,-1). c=-13. Therefore the axis of the hyperbola is y=4x-13. The centre or origin of the hyperbola lies midway between them at ((1/2)(5+3),(1/2)(7-1))=(4,3) and the equation of the conjugate axis is given by y=-x/4+k where k is found by plugging in the midpoint: 3=-4/4+k, so k=4 and y=4-x/4 is the equation of the conjugate axis. Let's define the axes of the hyperbola as X and Y instead of x and y. We can place the centre of the hyperbola at (0,0) in the X-Y plane so that in this plane the equation of the hyperbola is Y^2/A^2-X^2/B^2=1. The vertices are found by finding Y when X=0, so (0,A) and (0,-A) are the vertices. The asymptotes are given by (Y/A+X/B)(Y/A-X/B)=0. So Y=AX/B and Y=-AX/B are their equations. The distance between the vertices is 2A so we can work this out from the x-y coordinates of the vertices using Pythagoras' Theorem: 2A=sqrt((7+1)^2+(5-3)^2)=sqrt(68). A=sqrt(68)/2, A^2=68/4=17. Since the hyperbola is rectangular, B=A, and the equation of the hyperbola is Y^2-X^2=17. The asymptotes then become Y=X and Y=-X. Now we have to transform X-Y to x-y, so that the equation can be expressed in the x-y plane. We use geometry and trigonometry  for this. Let O'(4,3) be the location of the origin of the X-Y plane. Let P(x,y) be any point. In the X-Y plane this is P'(X,Y). Join O'P'=O'P. This line can be represented in both reference frames. O'P^2=X^2+Y^2=(x-4)^2+(y-3)^2 by Pythagoras. The gradient of the Y axis with respect to the x-y frame is 4, and we can represent this as tan(w) so w=tan^-1(4) or tan(w)=4; sin(w)=4/sqrt(17); cos(w)=1/sqrt(17). The angle z that O'P makes with the line y=3 is given by tan(z)=(y-3)/(x-4). sin(z)=(y-3)/O'P; cos(z)=(x-4)/O'P. Let v=90-w. sin(v+z)=Y/O'P; cos(v+z)=X/O'P. v+z=90-w+z=90-(w-z), so sin(v+z)=cos(w-z)=cos(w)cos(z)+sin(w)sin(z)=(x-4)/(O'Psqrt(17))+4(y-3)/(O'Psqrt(17))=(x+4y-16)/(O'Psqrt(17)). cos(v+z)=sin(w-z)=sin(w)cos(z)-cos(w)sin(z)=4(x-4)/(O'Psqrt(17))-(y-3)/(O'Psqrt(17))=(4x-y-13)/(O'Psqrt(17)). Y=O'P*sin(v+z)=(x+4y-16)/sqrt(17); X=O'P*cos(v+z)=(4x-y-13)/sqrt(17). So P maps to the x-y plane using these transformations. Y^2-X^2=17 becomes (x+4y-16)^2-(4x-y-13)^2=289, the equation of the hyperbola. This expands to x^2+8x(y-4)+16y^2-128y+256-(16x^2-8x(y+13)+y^2+26y+169)=289; -15x^2+16xy+15y^2+72x-154y-202=0. The asymptotes are x+4y-16+4x-y-13=0; 5x+3y-29=0 and x+4y-16-4x+y+13=0; 5y-3x-3=0.
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find the equation of the hyperbola one of its asymptotes has the equation 4x-2y+5 = 0 one vertex is v (-1, -1/2) and its conjugate axis is perpendicular to the y-axis.

The general equation of a hyperbola is (x-h)^2/a^2-(y-k)^2/b^2=+1. The asymptotes are the solutions of (x-h)^2/a^2-(y-k)^2/b^2=0, ((x-h)/a+(y-k)/b)((x-h)/a-(y-k)/b)=0. For +1 the hyperbola is horizontal, so are its foci, and for -1 it's vertical, so are its foci. The line joining the foci is perpendicular to the conjugate axis, which is parallel to the x axis in this case (perpendicular to the y axis), so the foci are located vertically in line, parallel to the y axis. The equation is therefore (x-h)^2/a^2-(y-k)^2/b^2=-1. (h,k) is the centre or origin of the hyperbola, and represents the displacement of the axes from (0,0). Whether the hyperbola is horizontal or vertical doesn't alter the asymptotes. So (x-h)/a+(y-k)/b)=4x-2y+5 is one of the asymptotes; or (x-h)/a-(y-k)/b)=4x-2y+5. This asymptote has a positive slope, so, since the asymptotes form an X, it's the / one. The other asymptote has negative slope \. Therefore, x/a=4x and y/b=2y, so a=1/4 and b=1/2. The constant 5=-4h+2k.  To find the vertices, put x=h, then y-k=+b=+1/2. The vertices are (h,k+1/2) and (h,k-1/2), having a common x coord, h. So h=-1, and k is given by 5=4+2k, k=1/2. Therefore, the equation of the hyperbola is 16(x+1)^2-4(y-1/2)^2=-1. This expands to 16x^2+32x-4y^2+4y+16=0. The \ asymptote has the equation 4(x+1)+2(y-1/2)=4x+2y+3=0. A graph of the hyperbola is shown below.
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find the asymptotes of the hyperbola

The asymptotes are straight lines with opposite gradient representing the limits for x and y as they approach infinity and minus infinity. The hyperbola is in standard form but the origin is offset from (0,0) to the point (-2,2). The asymptotes cross at this point like the centre of an X. The slopes of the asymptotes are found by taking the square roots of the x and y terms: (x+2)/20 and (y-2)/15. When x and, therefore, y are very large we have (x+2)/20=+/-(y-2)/15, so y=2+/-15/20, i.e., 2+/-3/4. Therefore the slopes of the asymptotes are 3/4 and -3/4. The standard form of a linear equation is y=ax+b, where a is the slope, so we can write for each asymptote y=3x/4+b and y=-3x/4+b, where b assumes a different value in each equation. The asymptotes cross at (-2,2) so these values of x and y satisfy both equations. This enables us to find b in each case. 2=3*(-2)/4+b and 2=-3*(-2)/4+b. This gives values 7/2 and 1/2 for b, so the asymptotes are y=3x/4+7/2 and y=1/2-3x/4.
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Trace the conicoid –(x^2/4) +..... (Analytical Geometry)

The picture below shows an aerial view of the conicoid as seen looking down the y axis at the x-z plane. The outer ellipse is the way the conicoid appears at a distance of 3sqrt(5) from the origin, while the inner one is distance 3sqrt(2) from the origin. The origin itself is the point (0,3,0) as seen from above. The straight line is the line z=x, which is the edge view of the plane z=x. The origin is where the vertex of the conicoid is. The two vertical lines on the left are just markers showing the extent of the ellipse's radius in the x direction. The ellipse's radius in the z direction is 8 for the outer ellipse and 4 for the inner. The picture clearly shows that the plane z=x intersects the conicoid. This view is also the view looking towards the origin from the negative side of y, because the conicoid is in fact two shapes which are reflections of each other. The origin is the point (0,-3,0). Now we adjust our viewpoint and look along the other axes to get an idea of other aspects of the conicoid. Start with the x-y plane, viewed along the z axis. If we treat z in the equation as a constant we can see what the curve looks like for x and y. Initially put z=0: -x^2/4+y^2/9=1 is the equation of a hyperbola, with asymptotes given by x^2/4=y^2/9 or y=±3x/2. These are represented by two lines y=3x/2 and -3x/2, which lie on intersecting planes. When z^2/16+1=4, -x^2/4+y^2/9=4 and the asymptotes are x^2/16=y^2/36, y=3x/2 and -3x/2 as before, and z=4sqrt(3). No matter where we are along the z axis the asymptotes are the same, so the conicoid is contained within these planes. The picture below shows the view along the z axis. In this picture the line y=x (edge-on view of the plane) misses the conicoid completely, because it lies outside the asymptotes. You can also see the hyperbolas for increasing values of z. The inner hyperbolas are when z=0 and the outer ones are when z=4sqrt(3). By joining one side of each hyperbola to the other with a horizontal line you can visualise the edge-on appearance of the ellipses. These are x-diameters of each ellipse. Finally, the view from the x axis. Here the line y=z lies inside the asymptotes so the plane intersects the conicoid.
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Find the vertices, co-vertices, and asymptotes of the hyperbola. X squared/16-y squared/36=1

(x/4)^2-(y/6)^2=1 is another way of writing the equation for the hyperbola. The asymptotes are given by y=±6x/4=±3x/2, i.e., the lines y=3x/2 and y=-3x/2. The vertices are where the curves cross the x axis (y=0), when x/4=±1, so x=-4 and 4, (-4,0) and (4,0).
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find the intercepts with the axes or asyptotes, and draw a sketch of: y=(6/x)

There are no intercepts, because the equation is not defined when either x or y is zero. However, the axes are asymptotes for the hyperbolic curve. For example, when x or y is 1 y or x is 6. For x or y is 12, y or x is 1/2. For x or y = 48, y or x is 1/8. So when x and y get large and positive, the curve comes closer to each axis. The same happens by reflection for negativity be values. Now the graph can be drawn. In the 1st quadrant and 3rd quadrants we can see the hyperbolas. There is no curve in quadrants 2 and 4. So the hyperbolas have a sort of L or an inverted L shape in quadrants 1 and 3, with a curve between 0 and 1, and 0 and -1, for x and y joining the vertical part of the L to the horizontal part.
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R(x)=x(x-20)^2/(x+8)^3

Domain of R: x<>-8 (x not equal to -8) because x+8 would be zero and division by zero is not permissible. x=-8 is equation of vertical asymptote. When x is very large and positive or negative, R(x) approaches x^3/x^3=1, so y=1 is the horizontal asymptote. Draw the asymptotes. On the left of x=-8 R(x) is large and positive and drops steeply to continue closer to the horizontal asymptote as x becomes more negative. On the right of x=-8 R(x) rises steeply from a large negative value passing through the origin and rising a little above the x axis but levelling out to approach the horizontal asymptote. At x=2.5 the curve rises to its local flattish maximum at R(x)=0.66 approx., dropping to zero at x=20, then it rises towards the asymptote gradually as x gets larger.
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Give the equation of the horizontal asymptote, if any, of the function f(x) = (x+3)/(x^2 -25)

This function has asymptotes at x=+/-5, because these values produce zero in the denominator. These, however, are  vertical asymptotes, lines parallel to the vertical axis f(x). f(x)=(x+3)/((x-5)(x+5)). As x approaches infinity the function approaches 1/x which approaches zero. So the horizontal x axis is also an asymptote.
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Vertical Asyn of (x+1) Horizontal of 3/4 and Y intercept of 2 what is the equation?


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coordinates of vertices and foci,also equation of directrix of the rectangular hyperbola xy=c^2

my answer is  xy=c^2=>y=c^2/x the gradient function is y=-c^2/x^2 the gradient,m is at the point(ct,c/t is m=-c^/(ct)^2=-1/+^2 the equation of tagnent.
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