Question

There are exactly two points on the ellipse x²/a²+y²/b²=1 whose distance from its centre is same and equal to (1/√2)(√a²+2b²) Then the eccentricity of the ellipse is

A. 1/2
B. 1/√2
C. √2/3
D. √3/2

Answer 1

To find the eccentricity of the ellipse, you can use the formula e = √(1 - (b^2/a^2)). Given the equation of the ellipse and the distance from the center to the points, you can solve for x² and y² and substitute them into the formula to find the eccentricity.

Step-by-step explanation:

The eccentricity of an ellipse can be determined by the formula:

e = √(1 - (b2/a2))

Given the equation of the ellipse: x²/a² + y²/b² = 1 and the distance from the center to the points on the ellipse is (1/√2)(√a²+2b²), we can calculate the eccentricity.

Let's assume P(x, y) is a point on the ellipse. The distance from P to the center (0, 0) is given by d = √(x² + y²).

Substituting the equation of the ellipse into the distance formula, we get:

d = √((x²(a² + b²) + b²(a² - x²))/(a²))

Since the distance from the center to the points on the ellipse is (1/√2)(√a²+2b²), we can equate it to d:

(1/√2)(√a² + 2b²) = √((x²(a² + b²) + b²(a² - x²))/(a²))

Squaring both sides of the equation and simplifying, we can solve for x² and y² in terms of a² and b².

Then we substitute these in the formula for eccentricity to find e.