The tangent and the normal lines at the point to the circle and the -axis form a triangle. The area of this triangle (in square units) is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
At (√3, 1), the normal to x² + y² = 4 passes through the origin (since it's a circle centered at origin), so the normal line is y = x/√3, which meets the x-axis at the origin.
The tangent at (√3, 1) has slope -√3 (negative reciprocal of normal slope 1/√3), giving equation: y - 1 = -√3(x - √3), i.e., y = -√3x + 4. This meets the x-axis at x = 4/√3.
The triangle has vertices at origin (0,0), (√3, 0) (foot of perpendicular from point to x-axis — actually the base is along x-axis from 0 to 4/√3), and (√3, 1).
Base = distance along x-axis from origin to (4/√3, 0) = 4/√3. Height = y-coordinate of (√3, 1) = 1.
Area = (1/2) × (4/√3) × 1 = 2/√3.
The tangent at (√3, 1) has slope -√3 (negative reciprocal of normal slope 1/√3), giving equation: y - 1 = -√3(x - √3), i.e., y = -√3x + 4. This meets the x-axis at x = 4/√3.
The triangle has vertices at origin (0,0), (√3, 0) (foot of perpendicular from point to x-axis — actually the base is along x-axis from 0 to 4/√3), and (√3, 1).
Base = distance along x-axis from origin to (4/√3, 0) = 4/√3. Height = y-coordinate of (√3, 1) = 1.
Area = (1/2) × (4/√3) × 1 = 2/√3.
[JEE Main 2019 · 8 Apr · Shift II]