Question

For \(\alpha\) belonging to an interval of length \(\beta\), suppose \((\alpha,-\alpha)\) is an interior point of the ellipse \[ 4x^2+5y^2=1. \] Then \[ (6\beta-4)^{201}+201= \]

• A. \(202\)
• B. \(0\)
• C. \(402\)
• D. \(201\)

Hint:

For an interior point of an ellipse, substitute the coordinates into the ellipse equation and use a strict inequality (\(<1\) after normalization) to determine the allowable interval.

Answer 1

The Correct Option is D

Step 1: Apply the interior point condition for the ellipse.
The ellipse is 4x² + 5y² = 1. For (α, -α) to lie inside, we need 4α² + 5α²<1 → 9α²<1 → α²<1/9 → -1/3<α<1/3.

Step 2: Compute the length of the admissible interval.
The length β = (1/3) - (-1/3) = 2/3.

Step 3: Evaluate the required expression.
6β - 4 = 6(2/3) - 4 = 4 - 4 = 0. Then (6β - 4)^201 + 201 = 0^201 + 201 = 201.

Step 4: Final conclusion.
The expression evaluates to 201.