The foci of a hyperbola are ( ± 2 , 0 ) and its eccentricity is 3 2 . A tangent, perpendicular to the line 2 x + 3 y = 6 , is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the x - and y -axes are a and b respectively, then | 6 a | + | 5 b | is equal to

Step-by-step solution

The given equation of hyperbola is x 2 a 2 - y 2 b 2 = 1 , a e = 2 , e = 3 2 ⇒ a = 4 3 ⇒ b 2 = a 2 e 2 - a 2 = 4 - 16 9 = 20 9 The tangent is perpendicular to 2 x + 3 y = 6 . Slope of tangent is m = - 1 - 2 3 = 3 2 Equation of tangent, m = 3 2 Equation of tangent is y = m x - a 2 m 2 - b 2 ⇒ y = 3 2 x - 16 9 × 9 4 - 20 9 ( ∵ Tangent is in 1 st  quadrant ⇒ C < 0 ⇒ y = 3 2 x - 4 3 Converting the equation into intercept form: ⇒ x 8 9 + y - 4 3 = 1 ⇒ 6 a + 5