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# If equation of the directrix of a hyperbola is2x+y-1=0 and eccentricity is √3 , focus of the hyperbola is (1,1) then find equation of the hyperbola.

Arun
25763 Points
2 years ago
Let P(x, y) is any point on the hyperbola.
given, focus of parabola is S(1,1).
equation of directrix is 2x + y = 1
From P draw PM perpendicular to the directrix then PM = (2x + y – 1)/√(2² + 1²) = (2x + y – 1)/√5
Also from the definition of the hyperbola, we have
SP/PM = e ⇒ SP = ePM
⇒ √{(x–1)² + (y–1)²} = √3{(2x + y – 1)/√5}
⇒ (x – 1)² + (y – 1)² = 3 (2x + y – 1)²/5
⇒ 5[(x² – 2x + 1) + (y² –2y + 1)] = 3(4x² + y² + 1 + 4xy – 4x – 2y)
⇒5x² - 10x + 5 + 5y² - 10y + 5 = 12x² + 3y² + 3 + 12xy - 12x - 6y
⇒7x² + 2y² + 12xy - 2x + 4y - 7 = 0
hence, equation of hyperbola is 7x² + 2y² + 12xy - 2x + 4y - 7 = 0