Class 12 Applications Of Derivatives · Very Long Answer5 Markstangent_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #76

The tangent to x²+y²=4 at a point in the open first quadrant meets the positive axes at A and B. With O the origin, find the minimum of OA+OB.

✅ Step-by-Step Solution

Final Answer: 4√2 units

Write the tangent point as P=(2cosθ,2sinθ), 0<θ<π/2. Its tangent is x cosθ+y sinθ=2.
The positive intercept lengths are OA=2secθ and OB=2cscθ, so F=2(secθ+cscθ).
F′=2(secθ tanθ−cscθ cotθ)=0 gives sin³θ=cos³θ and hence θ=π/4.
F′ has the sign of sin³θ−cos³θ: negative before π/4 and positive after it. Also F tends to infinity at either endpoint, proving the global minimum.
Fmin=2(√2+√2)=4√2 units.

[1 mark] Write the tangent point as P=(2cosθ,2sinθ), 0<θ<π/2. Its tangent is x cosθ+y sinθ=2.
[1 mark] The positive intercept lengths are OA=2secθ and OB=2cscθ, so F=2(secθ+cscθ).
[1 mark] F′=2(secθ tanθ−cscθ cotθ)=0 gives sin³θ=cos³θ and hence θ=π/4.
[1 mark] F′ has the sign of sin³θ−cos³θ: negative before π/4 and positive after it. Also F tends to infinity at either endpoint, proving the global minimum.
[1 mark] Fmin=2(√2+√2)=4√2 units.

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