If the maximum distance of normal to the ellipse x²/4+y²/b²=1, b < 2, from the origin is 1, then the eccentricity of the ellipse is :

Solution

Maths sir · 4 steps

Write the normal at a general point, find its distance from O, then push the denominator as low as it goes.

1. Normal at a point. Take a = 2 and the point (2 cos θ, b sin θ). The normal there is ax sec θ - by cosec θ = a² - b².
2. Distance from origin. Use the perpendicular distance formula with the point (0, 0). The numerator a² - b² is fixed, so the distance is largest when the denominator is smallest.

   d = (a² - b²)/√(a²sec² θ + b² cosec²θ)

3. Minimise the denominator. a²sec² θ + b² cosec²θ = a² + b² + a²tan² θ + b²cot² θ ≥ a² + b² + 2ab = (a+b)² by AM-GM. So dₘₐₓ = (a² - b²)/(a + b) = a - b.
4. So. 2 - b = 1 gives b = 1. Then e = √(1 - b²/a²) = √(1 - 1/4) = √3/2, which is option (d).

The idea to keep. For an ellipse, the farthest any normal gets from the centre is a - b. Remember it, and questions like this one take a single line.

Answer from the NTA final answer key. Solution by PrepWiser.