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
\[
(6\beta-4)^{201}+201=
\]
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
\[
(6\beta-4)^{201}+201=
\]
Show 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.
Updated On: Jun 18, 2026
- \(202\)
- \(0\)
- \(402\)
- \(201\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use the condition that \((\alpha,-\alpha)\) lies inside the ellipse.
The ellipse is \[ 4x^2+5y^2=1. \] Since \((\alpha,-\alpha)\) is an interior point, \[ 4\alpha^2+5(-\alpha)^2<1. \] \[ 4\alpha^2+5\alpha^2<1. \] \[ 9\alpha^2<1. \] \[ \alpha^2<\frac{1}{9}. \] \[ -\frac13<\alpha<\frac13. \]
Step 2: Find the length of the interval.
Thus \(\alpha\) belongs to the interval \[ \left(-\frac13,\frac13\right). \] Hence the length of the interval is \[ \beta=\frac13-\left(-\frac13\right) =\frac23. \]
Step 3: Evaluate \(6\beta-4\).
\[ 6\beta-4 = 6\left(\frac23\right)-4. \] \[ =4-4. \] \[ =0. \]
Step 4: Evaluate the given expression.
\[ (6\beta-4)^{201}+201 = 0^{201}+201. \] \[ =201. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{201} \]
The ellipse is \[ 4x^2+5y^2=1. \] Since \((\alpha,-\alpha)\) is an interior point, \[ 4\alpha^2+5(-\alpha)^2<1. \] \[ 4\alpha^2+5\alpha^2<1. \] \[ 9\alpha^2<1. \] \[ \alpha^2<\frac{1}{9}. \] \[ -\frac13<\alpha<\frac13. \]
Step 2: Find the length of the interval.
Thus \(\alpha\) belongs to the interval \[ \left(-\frac13,\frac13\right). \] Hence the length of the interval is \[ \beta=\frac13-\left(-\frac13\right) =\frac23. \]
Step 3: Evaluate \(6\beta-4\).
\[ 6\beta-4 = 6\left(\frac23\right)-4. \] \[ =4-4. \] \[ =0. \]
Step 4: Evaluate the given expression.
\[ (6\beta-4)^{201}+201 = 0^{201}+201. \] \[ =201. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{201} \]
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