Let be the point of intersection of the common tangents to the parabola and the hyperbola . If and denotes the foci of the hyperbola where lies on the positive -axis then divides in a ratio
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
The common tangent to y² = 12x has the form y = mx + 3/m (using tangent condition c = a/m for parabola y² = 4ax with a = 3).
For this line to be tangent to 8x² - y² = 8 (i.e., x²/1 - y²/8 = 1), the condition is c² = a²m² - b², giving (3/m)² = 1·m² - 8, so 9/m² = m² - 8, leading to m⁴ - 8m² - 9 = 0, i.e., (m² - 9)(m² + 1) = 0, so m = ±3.
The two tangents are y = 3x + 1 and y = -3x - 1. Their intersection: 3x + 1 = -3x - 1 → 6x = -2 → x = -1/3, y = 0. So P = (-1/3, 0).
For the hyperbola x²/1 - y²/8 = 1: a² = 1, b² = 8, so c² = 9, c = 3. The foci are S = (3, 0) and S' = (-3, 0) (S on positive x-axis).
P = (-1/3, 0) divides S(3,0) and S'(-3,0): using section formula, if P divides SS' in ratio k:1, then (-1/3) = (k·(-3) + 1·3)/(k+1) → -k/
For this line to be tangent to 8x² - y² = 8 (i.e., x²/1 - y²/8 = 1), the condition is c² = a²m² - b², giving (3/m)² = 1·m² - 8, so 9/m² = m² - 8, leading to m⁴ - 8m² - 9 = 0, i.e., (m² - 9)(m² + 1) = 0, so m = ±3.
The two tangents are y = 3x + 1 and y = -3x - 1. Their intersection: 3x + 1 = -3x - 1 → 6x = -2 → x = -1/3, y = 0. So P = (-1/3, 0).
For the hyperbola x²/1 - y²/8 = 1: a² = 1, b² = 8, so c² = 9, c = 3. The foci are S = (3, 0) and S' = (-3, 0) (S on positive x-axis).
P = (-1/3, 0) divides S(3,0) and S'(-3,0): using section formula, if P divides SS' in ratio k:1, then (-1/3) = (k·(-3) + 1·3)/(k+1) → -k/
[JEE Main 2019 · 12 Apr · Shift I]