The equation of the normal to the ellipse x24+y22=1 at the point whose eccentric angle is π/4 is

a2=4 b2=2 θ=π4 normal equation axcosθ-bysinθ=a2-b2 2x12-2y12=4-2 22x-2y=2

The equation of the normal to the ellipse x24+y22=1 at the point whose eccentric angle is π/4 is

a2=4 b2=2 θ=π4 normal equation axcosθ-bysinθ=a2-b2 2x12-2y12=4-2 22x-2y=2

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The equation of the normal to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1$ at the point whose eccentric angle is $π / 4$ is

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$x + \sqrt{2} y = 2 \sqrt{2}$

$\sqrt{2} x - y = 1$

$x - \sqrt{2} y = 0$

$\sqrt{2} x + y = 3$

answer is B.

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Detailed Solution

$a^{2} = 4 b^{2} = 2 θ = \frac{π}{4} normal equation \frac{ax}{cosθ} - \frac{by}{sinθ} = a^{2} - b^{2} \frac{2 x}{\frac{1}{\sqrt{2}}} - \frac{\sqrt{2} y}{\frac{1}{\sqrt{2}}} = 4 - 2 2 \sqrt{2} x - 2 y = 2$

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