If the line , touches both the curves and , then is equal to
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
For the line ax + y = c (i.e., y = c - ax) to touch y² = 4√2·x, use the tangency condition for y² = 4ax: the line y = mx + c touches the parabola when c = a/m. Here the parabola has parameter a = √2, and the line has slope m = -a (the coefficient of x). So tangency requires c = √2/(-a) ... but we need the condition properly.
For y² = 4√2·x, a tangent line y = mx + k satisfies k = √2/m. With y = -ax + c, slope m = -a, so c = √2/(-a) = -√2/a.
For the line ax + y = c to touch x² + y² = 1, the distance from origin to the line equals 1:
|c|/√(a² + 1) = 1, so c² = a² + 1.
From the parabola condition: c = -√2/a, so c² = 2/a².
Setting equal: 2/a² = a² + 1 → a⁴ + a² - 2 = 0 → (a² + 2)(a² - 1) = 0 → a² = 1, so a = ±1.
Then c² = 2/1 = 2, giving |c| = √2.
For y² = 4√2·x, a tangent line y = mx + k satisfies k = √2/m. With y = -ax + c, slope m = -a, so c = √2/(-a) = -√2/a.
For the line ax + y = c to touch x² + y² = 1, the distance from origin to the line equals 1:
|c|/√(a² + 1) = 1, so c² = a² + 1.
From the parabola condition: c = -√2/a, so c² = 2/a².
Setting equal: 2/a² = a² + 1 → a⁴ + a² - 2 = 0 → (a² + 2)(a² - 1) = 0 → a² = 1, so a = ±1.
Then c² = 2/1 = 2, giving |c| = √2.
[JEE Main 2019 · 10 April · Shift II]