Equation of a common tangent to the circle, and the parabola, , is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
The tangent to y² = 4x in slope form is y = mx + 1/m. For this line to be tangent to the circle (centre (3,0), radius 3), the distance from (3,0) to the line mx - y + 1/m = 0 must equal 3.
Distance = |3m - 0 + 1/m| / √(m² + 1) = 3
So (3m + 1/m)² = 9(m² + 1)
9m² + 6 + 1/m² = 9m² + 9
1/m² = 3, so m² = 1/3, m = ±1/√3.
Taking m = 1/√3: y = x/√3 + √3, multiply through by √3: √3·y = x + 3. This matches option 3.
Distance = |3m - 0 + 1/m| / √(m² + 1) = 3
So (3m + 1/m)² = 9(m² + 1)
9m² + 6 + 1/m² = 9m² + 9
1/m² = 3, so m² = 1/3, m = ±1/√3.
Taking m = 1/√3: y = x/√3 + √3, multiply through by √3: √3·y = x + 3. This matches option 3.
[JEE Main 2019 · 9 Jan · Shift I]