Consider the ellipse with the equation $x^{2} + 3 y^{2} - 2 x - 6 y - 2 = 0.$ The eccentric angle of a point on the ellipse at a distance 2 units from the contra of the ellipse is

\frac{π}{4}

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\frac{π}{2}

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\frac{π}{6}

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\frac{π}{3}

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Solution

The correct option is A $\frac{π}{4}$
We have, $x^{2} + 3 y^{2} - 2 x - 6 y - 2 = 0.$

$\Rightarrow$ $x^{2} - 2 x + 1 - 1 + 3 y^{2} - 6 y + 3 - 3 - 2 = 0.$

$\Rightarrow$ $(x - 1)^{2} + 3 (y - 1)^{2} = 6$

$\Rightarrow$ $(x - 1)^{2} + 3 (y - 1)^{2} = 6$ which becomes $x^{2} + 3 y^{2} = 6$ on shifting the origin to $(1, 1)$ . Any point with eccentric angle $θ$ is $(\sqrt{6} c o s θ, \sqrt{2} s i n θ)$
$\Rightarrow$ $4 = 6 c o s^{2} θ + 2 s i n^{2} θ \Rightarrow 4 c o s^{2} θ = 2 \Rightarrow c o s θ = \pm \frac{1}{\sqrt{2}}$
$\Rightarrow$ Hence $θ = \frac{π}{4}$

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A. If the tangent to the ellipse $x^{2} + 4 y^{2} = 16$ at the point $P (ϕ)$ is a normal to the circle $x^{2} + y^{2} - 8 x - 4 y = 0$ , then $\frac{ϕ}{2}$ may be	1. $0$
B. The eccentric angle(s) of a point on the ellipse $x^{2} + 3 y^{2} = 6$ at a distance $2$ units from the centre of the ellipse is/are	2. ${cos}^{- 1} (- \frac{2}{3})$
C. The eccentric angle of intersection of the ellipse $x^{2} + 4 y^{2} = 4$ and the parabola $x^{2} + 1 = y$ is	3. $\frac{π}{4}$
D. If the normal at the point $P (θ)$ to the ellipse $\frac{x^{2}}{14} + \frac{y^{2}}{5} = 1$ intersects it again at the point $Q (2 θ)$ , then $θ$ is	4. $\frac{5 π}{4}$

θ, θ + \frac{π}{2}, θ + π, θ + \frac{3 π}{2}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

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