If the maximum distance of normal to the ellipse x 2 4 + y 2 b 2 = 1 , b 2 , from the origin is 1, then the eccentricity of the ellipse is: [2023]

Question

If the maximum distance of normal to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{b^{2}} = 1, b < 2,$ from the origin is 1, then the eccentricity of the ellipse is: [2023]

Smart Achievers · Accepted Answer

(2)

We have, $\frac{x^{2}}{2^{2}} + \frac{y^{2}}{b^{2}} = 1; b < 2$

Equation of normal at $(2 \cos θ, b \sin θ)$ is

$2 x \sec θ - b y c o s e c θ = 4 - b^{2} .$

According to the question, $| \frac{4 - b^{2}}{\sqrt{4 \sec^{2} θ + b^{2} {cosec}^{2} θ}} |_{\max}$ $= 1$

For maximum distance, $4 \sec^{2} θ + b^{2} {cosec}^{2} θ$ should be minimum.

$∴$ $\min (a^{2} \sec^{2} θ + b^{2} {cosec}^{2} θ) = {(a + b)}^{2}$

$∴$ $\Rightarrow 1 = \frac{4 - b^{2}}{\sqrt{{(2 + b)}^{2}}} \Rightarrow 1 = \frac{4 - b^{2}}{2 + b}$

$\Rightarrow 2 + b = 4 - b^{2} \Rightarrow b^{2} + b - 2 = 0$

$\Rightarrow b^{2} + 2 b - b - 2 = 0 \Rightarrow b (b + 2) - 1 (b + 2) = 0$

$\Rightarrow (b + 2) (b - 1) = 0 (∵ b < 2)$

$\Rightarrow b = 1$

Now, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$

Solution

Q.
If the maximum distance of normal to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{b^{2}} = 1, b < 2,$ from the origin is 1, then the eccentricity of the ellipse is: [2023]

1 $\frac{\sqrt{3}}{4}$

2 $\frac{\sqrt{3}}{2}$

3 $\frac{1}{\sqrt{2}}$

4 $\frac{1}{2}$

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Solution

Q. If the maximum distance of normal to the ellipse x24+y2b2=1,b<2, from the origin is 1, then the eccentricity of the ellipse is: [2023]

1 34

2 32

3 12