Question

Find the equation of hyperbola whose directrix is 2x+y=1,focus(1,1) and eccentricity root3

Answer 1

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