Let a tangent be drawn to the ellipse x²/27 + y² = 1 at (3√3 cos θ ,sin θ ) where 0 ∈ ( 0,π/2 ). Then the value of θ such that the sum of intercepts on axes made by this tangent is minimum is equal to :

Solution

Maths sir · 4 steps

Write the tangent in parametric form, read off both intercepts, then differentiate their sum in θ.

1. Tangent at the point. For x²/a² + y²/b² = 1 the tangent at (a cos θ, b sin θ) is (x cos θ)/a + (y sin θ)/b = 1. Here a = 3√3 and b = 1, so the tangent is (x cos θ)/3√3 + y sin θ = 1.
2. Read the intercepts. Put y = 0 to get the x-intercept 3√3 sec θ. Put x = 0 to get the y-intercept csc θ. Both are positive for θ ∈ (0, π/2), so the sum is f(θ) = 3√3 sec θ + csc θ.

   f'(θ) = 3√3 sec θ tan θ - csc θ cot θ = 0 ⇒ tan³ θ = 1/3√3 = (1/√3)³

3. Find the minimum. Writing the derivative in sines and cosines gives 3√3 sin³ θ = cos³ θ, so tan θ = 1/√3 and θ = π/6. Since f → ∞ at both ends of the interval and this is the only critical point, it is the minimum.

   θ	x-intercept 3√3 sec θ	y-intercept csc θ	Sum
   π/6	6	2	8
   π/8	≈ 5.62	≈ 2.61	≈ 8.24
   π/4	≈ 7.35	≈ 1.41	≈ 8.76
   π/3	≈ 10.39	≈ 1.15	≈ 11.55

4. So. The sum of intercepts is least, equal to 8, at θ = π/6. That is option (a).

The idea to keep. For a tangent to an ellipse, the parametric form gives intercepts a sec θ and b csc θ. Turn any intercept condition into a function of θ and use calculus.

Answer from the NTA final answer key. Solution by PrepWiser.