If the maximum distance of normal of the ellipse , from the origin is , then the eccentricity of the ellipse is :

Explanation: Solution Ellipse ki equation di gayi hai: Yahan (yani ) aur hai. Step 1: Equation of Normal Ellipse par kisi point par normal ki equation hoti hai: Values rakhne par: Step 2: Distance from Origin Origin se is line ki distance ( ) ka formula: Step 3: Maximum Distance Distance ko maximum karne ke liye niche wali term (denominator) minimum honi chahiye. Maximum distance humein di gayi hai: Yahan se: Step 4: Eccentricity ( ) Eccentricity ka formula jab ho: Correct Option: (3)

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If the maximum distance of normal of 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 :

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