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A tangent to the ellipse x²/27 + y² = 1 is taken at the point (√3 cos θ, sin θ), where θ lies in (0, π/2). For which value of θ will the sum of the intercepts cut off by this tangent on the coordinate axes be the least?
- π/3
- π/6
- π/8
- π/4
Correct answer: π/6
Solution
Tangent at (3sqrt3 cos th, sin th) has x-intercept 27/(3sqrt3 cos th) and y-intercept 1/sin th, so the sum S=3sqrt3/cos th + 1/sin th. dS/dth=0 gives tan^3 th=1/(3sqrt3), i.e. tan th=1/sqrt3, so th=pi/6 (minimum S=8).
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