If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :

Solution

Maths sir · 3 steps

Turn the given condition into one equation in b, a and e, then use b² = a²(1-e²).

1. Write the condition. For x²/a²+y²/b²=1 with a > b, the minor axis has length 2b and the foci (± ae, 0) are 2ae apart. The condition says 2b = 1/2(2ae), so b = ae/2.
2. Use the ellipse relation. Every ellipse satisfies b² = a²(1-e²). Put b = ae/2 into it and cancel a².

   e²/4 = 1 - e² ⇒ 5e²/4 = 1 ⇒ e² = 4/5

3. So. Eccentricity is positive, so e = 2/√5. This is option (b).

The idea to keep. Any condition linking the axes and the foci becomes an equation in a, b and e. Substitute it into b² = a²(1-e²), cancel a², and you are left with e alone.

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