JEE Mathematics 2024 Question with Solution

Given: Let $a$ be the semi-major axis, $b$ the semi-minor axis, and $2c$ the distance between the foci of the ellipse.

Find: The eccentricity $e$ of the ellipse.

For an ellipse,

$$e = \frac{c}{a}$$

and

$$b = a\sqrt{1 - e^2}$$

Since the length of the minor axis is equal to half of the distance between the foci,

$$2b = \frac{1}{2} \times 2c$$

so,

$$2b = c$$

Substitute $c = ae$:

$$2b = ae$$

Now substitute $b = a\sqrt{1 - e^2}$:

$$2a\sqrt{1 - e^2} = ae$$

Dividing by $a$,

$$2\sqrt{1 - e^2} = e$$

Squaring both sides,

$$4(1 - e^2) = e^2$$ $$4 = 5e^2$$ $$e^2 = \frac{4}{5}$$

Therefore,

$$e = \frac{2}{\sqrt{5}}$$

The correct option is D. The solution incorrectly labels the option as C, but the worked value matches option D.

Using the alternate relation

$$c^2 = a^2 - b^2$$

and from $2b = c$, we get

$$b = \frac{c}{2}$$

Substituting,

$$c^2 = a^2 - \left(\frac{c}{2}\right)^2$$ $$4c^2 = 4a^2 - c^2$$ $$5c^2 = 4a^2$$

Hence,

$$a^2 = \frac{5}{4}c^2$$

and so

$$e = \frac{c}{a} = \frac{2}{\sqrt{5}}$$

This again confirms option D.

Given: The minor axis length is half of the distance between the foci.

Find: Eccentricity of the ellipse.

The distance between the foci is $2c$ and the minor axis length is $2b$. Hence,

$$2b = \frac{1}{2}(2c)$$ $$2b = c$$ $$b = \frac{c}{2}$$

Now use the standard ellipse identity

$$c^2 = a^2 - b^2$$

Substitute $b = \frac{c}{2}$:

$$c^2 = a^2 - \frac{c^2}{4}$$ $$a^2 = c^2 + \frac{c^2}{4}$$ $$a^2 = \frac{5c^2}{4}$$

Therefore,

$$a = \frac{\sqrt{5}}{2}c$$

Now,

$$e = \frac{c}{a} = \frac{c}{(\sqrt{5}/2)c} = \frac{2}{\sqrt{5}}$$

Therefore, the eccentricity is $\frac{2}{\sqrt{5}}$, so the correct option is D.