SATHEE: Ellipse 2 Question 14

Ellipse 2 Question 14

14. Tangent is drawn to ellipse $\frac{x^{2}}{27} + y^{2} = 1$ at (3 $\sqrt{3} \cos θ, \sin θ)$ (where, $θ \in (0, π / 2)$ .

Then, the value of $θ$ such that the sum of intercepts on axes made by this tangent is minimum, is

$(2003, 1 M)$

(a) $\frac{π}{3}$

(b) $\frac{π}{6}$

(d) $\frac{π}{4}$

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Answer:

Correct Answer: 14. (b)

Solution:

Given, tangent is drawn at $(3 \sqrt{3} \cos θ, \sin θ)$ to $\frac{x^{2}}{27} + \frac{y^{2}}{1} = 1$ .

$∴$ Equation of tangent is $\frac{x \cos θ}{3 \sqrt{3}} + \frac{y \sin θ}{1} = 1$ .

Thus, sum of intercepts $= \frac{3 \sqrt{3}}{\cos θ} + \frac{1}{\sin θ} = f (θ)$ [say]

$\begin{aligned} \Rightarrow f^{'} (θ) = \frac{3 \sqrt{3} \sin^{3} θ - \cos^{3} θ}{\sin^{2} θ \cos^{2} θ}, put f^{'} (θ) = 0 \\ \Rightarrow \sin^{3} θ = \frac{1}{3^{3 / 2}} \cos^{3} θ \\ \Rightarrow \tan θ = \frac{1}{\sqrt{3}}, i.e. θ = \frac{π}{6} and at θ = \frac{π}{6}, f^{''} (0) > 0 \end{aligned}$

Hence, tangent is minimum at $θ = \frac{π}{6}$ .

Ellipse 2 Question 14

14. Tangent is drawn to ellipse $\frac{x^{2}}{27} + y^{2} = 1$ at (3 $\sqrt{3} \cos θ, \sin θ)$ (where, $θ \in (0, π / 2)$ .

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Ellipse 2 Question 14

14. Tangent is drawn to ellipse x227+y2=1 at (3 3cos⁡θ,sin⁡θ) (where, θ∈(0,π/2).

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14. Tangent is drawn to ellipse $\frac{x^{2}}{27} + y^{2} = 1$ at (3 $\sqrt{3} \cos θ, \sin θ)$ (where, $θ \in (0, π / 2)$ .