Given two integers A and B, representing the length of the semi-major and semi-minor axis of an ellipse with the equation (x2 / A2) + (y2 / B2) = 1, the task is to find the minimum area of the triangle formed by any tangent to the ellipse with the coordinate axes.
Input: A = 1, B = 2
Input: A = 2, B = 3
The idea is based on the observation that the equation of a tangent at coordinate (A * cosθ, B * sinθ) on the ellipse in the figure shown above is:
(x * cosθ / A) + (y * sinθ / B) = 1
The coordinates of P and Q are (A / cosθ, 0) and (0, B / sinθ) respectively.
Area of triangle formed by the tangent to the ellipse and the coordinate axes is given by:
Area of ΔOPQ = 0.5 * base * height = 0.5 * (A / cosθ) * (B / sinθ) = (A * B) / (2 * sinθ * cosθ) = (A * B) / sin2θ
Therefore, Area = (A * B) / sin2θ — (1)
To minimize the area, the value of sin2θ should be maximum possible, i.e. 1.
Therefore, the minimum area possible is A * B.
Below is the implementation of the above approach:
Time Complexity: O(1)
Auxiliary Space: O(1)
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