answer for (x+5)^2/11^2-(y+16)^2/6^2=1
answer for that Question is (-5,-16)
To write the generic equation of a hyperbola suggested by a graph, first identify the orientation of the hyperbola (horizontal or vertical) based on its shape. The standard forms are ((x - h)^2/a^2 - (y - k)^2/b^2 = 1) for a horizontal hyperbola and ((y - k)^2/a^2 - (x - h)^2/b^2 = 1) for a vertical hyperbola, where ((h, k)) is the center, and (a) and (b) are distances that determine the shape. Use points on the graph to find the values of (h), (k), (a), and (b) to complete the equation.
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The equation provided appears to have a typographical error, as it should likely be in the form of a standard circle equation. If you meant (x^2 + y^2 = 16), the center of the circle is at the coordinates (0, 0). If this is not the correct interpretation, please clarify the equation for an accurate response.
The transverse axis is a connection on a hyperbola. It connects the focus, or center, of the hyperbola, and can connect two together.
The equation of the circle is given by ((x - 6)^2 + (y + 5)^2 = 152). The general form of a circle's equation is ((x - h)^2 + (y - k)^2 = r^2), where ((h, k)) is the center and (r) is the radius. From the equation, the coordinates of the center of the circle are ((6, -5)).
Answers provided by: apexvs.com (x+26)2 + (y-11)2 = 1 _____ _____ 732 7 (-26, 11) or 4
The length of the latus rectum of a hyperbola is given by the formula ( \frac{2b^2}{a} ), where ( a ) is the distance from the center to the vertices and ( b ) is the distance from the center to the co-vertices. This length represents the width of the hyperbola at the points where it intersects the corresponding directrices. For hyperbolas oriented along the x-axis or y-axis, this formula applies similarly, with the values of ( a ) and ( b ) depending on the specific equation of the hyperbola.
The axes of the hyperbola.
To write the generic equation of a hyperbola suggested by a graph, first identify the orientation of the hyperbola (horizontal or vertical) based on its shape. The standard forms are ((x - h)^2/a^2 - (y - k)^2/b^2 = 1) for a horizontal hyperbola and ((y - k)^2/a^2 - (x - h)^2/b^2 = 1) for a vertical hyperbola, where ((h, k)) is the center, and (a) and (b) are distances that determine the shape. Use points on the graph to find the values of (h), (k), (a), and (b) to complete the equation.
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The equation provided appears to have a typographical error, as it should likely be in the form of a standard circle equation. If you meant (x^2 + y^2 = 16), the center of the circle is at the coordinates (0, 0). If this is not the correct interpretation, please clarify the equation for an accurate response.
Asymptotes are the guidelines that a hyperbola follows. They form an X and the hyperbola always gets closer to them but never touches them. If the transverse axis of your hyperbola is horizontal, the slopes of your asymptotes are + or - b/a. If the transverse axis is vertical, the slopes are + or - a/b. The center of a hyperbola is (h,k). I don't know what the rest of your questions are, though.
The transverse axis is a connection on a hyperbola. It connects the focus, or center, of the hyperbola, and can connect two together.
The equation of the circle is given by ((x - 6)^2 + (y + 5)^2 = 152). The general form of a circle's equation is ((x - h)^2 + (y - k)^2 = r^2), where ((h, k)) is the center and (r) is the radius. From the equation, the coordinates of the center of the circle are ((6, -5)).
To determine the center of a circle, you typically need the equation of the circle, which is usually in the form ((x - h)^2 + (y - k)^2 = r^2), where ((h, k)) represents the center coordinates and (r) is the radius. If you have specific coordinates or an equation for the circle labeled as "Imported Asset," please provide that information for a more accurate answer. Otherwise, the center is found at the point ((h, k)) derived from the equation.
Center
The general equation for the circle - or one of them - is: (x - a)^2 + (y - b)^2 = r^2 Where: a and b are the coordinates of the center r is the radius