Question

Find the equation of the given conic. horizantal ellipse with centre (5,1), major diameter 10, minor diameter 8

Answers

  1. The equation of the given conic, horizontal ellipse is (x – 5)²/5² + (y – 1)²/4² = 1 or 16x² + 25y² – 160x – 50y + 325 = 0.
    In mathematics, an ellipse is the location of points in a plane so that their separation from a fixed point has a set ratio of “e” to their separation from a fixed line (less than 1). The conic section, which is the intersection of a cone with a plane that does not intersect the base of the cone, includes the ellipse. The eccentricity of the ellipse is symbolized by the constant ratio “e,” the fixed point is known as the focus, and the fixed line is known as the directrix (d).
    The ellipse formula equation aids in the algebraic representation of an ellipse. You may use the following formula to determine an ellipse’s equation:
    The ellipse’s equation with a center at (0, 0) is as follows: x²/a² + y²/b² = 1.
    An ellipse with a center at (h,k) has the following equation: (x – h)²/a² + (y – k)²/(b²) = 1, where a and b are the semi-major and semi-minor diameters.
    In the question, we are given that the center (h,k) = (5,1), the major diameter = 10, or, a = 10/2 = 5, and the minor diameter = 8, or, b = 8/2 = 4.
    Substituting in the equation, we get:
    (x – 5)²/5² + (y – 1)²/4² = 1, in the general form, and,
    16(x² + 25 – 10x) + 25(y² + 1 – 2y) = 100,
    or, 16x² + 400 – 160x + 25y² + 25 – 50y = 100,
    or, 16x² + 25y² – 160x – 50y + 325 = 0.
    Thus, the equation of the given conic, horizontal ellipse is (x – 5)²/5² + (y – 1)²/4² = 1 or 16x² + 25y² – 160x – 50y + 325 = 0.
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