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 

• A. $\frac{\sqrt{3}}{4}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}$

Hint:

For such problems, we solve by simplifying the given equations step by step and applying geometrical concepts like normal distances and eccentricity.

Answer 1

The Correct Option is B

Equation of normal is
$2x\sec \theta -\text{by}\mathrm{cosec}\theta =4-b^2$
Distance from $(0,0)=\frac{4-b^2}{\sqrt{4{\sec }^2\theta +b^2{\mathrm{cosec}}^2\theta }}$
Distance is maximum if
$4{\sec }^2\theta +b^2{\mathrm{cosec}}^2\theta$ is minimum
$\Rightarrow {\tan }^2\theta =\frac{b}{2}$
$\Rightarrow \frac{4-b^2}{\sqrt{4\cdot \frac{b+2}{2}+b^2\cdot \frac{b+2}{b}}}=1$
$\Rightarrow 4-b^2=b+2\Rightarrow b=1\Rightarrow e=\frac{\sqrt{3}}{2}$

Answer 2

The given equation of the ellipse is:

\( \frac{x^2}{4} + \frac{y^2}{b^2} = 1 \) where \( b < 2 \).

The maximum normal distance from the origin is given by:

\( \frac{a^2 + b^2}{\sqrt{a^2 + b^2 - c^2}} = 1 \).

Substituting \( a^2 = 4 \) and using \( c^2 = a^2 - b^2 \), we solve for \( b \).

After simplifications, we find \( b = 1 \).

Now, eccentricity \( e = \frac{c}{a} = \frac{\sqrt{3}}{2} \).

Final Answer: √3/2.