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Tangents are drawn to the ellipse x² + 2y² = 2 at every point except its four vertices. The midpoints of the segments of these tangents cut off between the coordinate axes lie on which curve?
- x²/2 + y²/4 = 1
- x²/4 + y²/2 = 1
- 1/(2x²) + 1/(4y²) = 1
- 1/(4x²) + 1/(2y²) = 1
Correct answer: 1/(2x²) + 1/(4y²) = 1
Solution
The tangent's intercepts give h = sqrt2/(2 cos t) and k = 1/(2 sin t); eliminating t with cos² + sin² = 1 yields 1/(2x²) + 1/(4y²) = 1.
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