Question

find the equation of the hyperbola with focus (2, 2), eccentricity 2 and directrix 3x-4y-1=0​

Answer 1

Answer:

(i) focus is (0, 3), directrix is x + y – 1 = 0 and eccentricity = 2 Given: Focus = (0, 3) Directrix => x + y – 1 = 0 Eccentricity = 2 Now, let us find the equation of the hyperbola Let ‘M’ be the point on directrix and P(x, y) be any point of the hyperbola. By using the formula, e = PF/PM PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix] So, [We know that (a – b)2 = a2 + b2 + 2ab &(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac] So, 2{x2 + y2 + 9 – 6y} = 4{x2 + y2 + 1 – 2x – 2y + 2xy} 2x2 + 2y2 + 18 – 12y = 4x2 + 4y2+ 4 – 8x – 8y + 8xy 2x2 + 2y2 + 18 – 12y – 4x2 – 4y2 – 4 – 8x + 8y – 8xy = 0 – 2x2 – 2y2 – 8x – 4y – 8xy + 14 = 0 -2(x2 + y2 – 4x + 2y + 4xy – 7) = 0 x2 + y2 – 4x + 2y + 4xy – 7 = 0 ∴The equation of hyperbola is x2 + y2 – 4x + 2y + 4xy – 7 = 0 (ii) focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2 Focus = (1, 1) Directrix => 3x + 4y + 8 = 0 Eccentricity = 2 Now, let us find the equation of the hyperbola Let ‘M’ be the point on directrix and P(x, y) be any point of the hyperbola. By using the formula, e = PF/PM PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix] So, [We know that (a – b)2 = a2 + b2 + 2ab &(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac] 25{x2 + 1 – 2x + y2 + 1 – 2y} = 4{9x2 + 16y2+ 64 + 24xy + 64y + 48x} 25x2 + 25 – 50x + 25y2 + 25 – 50y = 36x2 + 64y2 + 256 + 96xy + 256y + 192x 25x2 + 25 – 50x + 25y2 + 25 – 50y – 36x2 – 64y2 – 256 – 96xy – 256y – 192x = 0 – 11x2 – 39y2 – 242x – 306y – 96xy – 206 = 0 11x2 + 96xy + 39y2 + 242x + 306y + 206 = 0 ∴The equation of hyperbola is11x2 + 96xy + 39y2 + 242x + 306y + 206 = 0 (iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity =√3 Given: Focus = (1, 1) Directrix => 2x + y = 1 Eccentricity =√3 Now, let us find the equation of the hyperbola Let ‘M’ be the point on directrix and P(x, y) be any point of the hyperbola. By using the formula, e = PF/PM PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix] So, [We know that (a – b)2 = a2 + b2 + 2ab &(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac] 5{x2 + 1 – 2x + y2 + 1 – 2y} = 3{4x2 + y2+ 1 + 4xy – 2y – 4x} 5x2 + 5 – 10x + 5y2 + 5 – 10y = 12x2 + 3y2 + 3 + 12xy – 6y – 12x 5x2 + 5 – 10x + 5y2 + 5 – 10y – 12x2 – 3y2 – 3 – 12xy + 6y + 12x = 0 – 7x2 + 2y2 + 2x – 4y – 12xy + 7 = 0 7x2 + 12xy – 2y2 – 2x + 4y– 7 = 0 ∴The equation of hyperbola is7x2 + 12xy – 2y2 – 2x + 4y– 7 = 0Read more on Sarthaks.com - https://www.sarthaks.com/805834/find-the-equation-of-the-hyperbola-whose-i-focus-is-0-3-directrix-is-x-y-1-0-and-eccentricity-2