Q.If $e$ is the eccentricity of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $(a < b)$, then
(A) $b^2 = a^2(1 - e^2)$
(B) $a^2 = b^2(1 - e^2)$
(C) $a^2 = b^2(e^2 - 1)$
(D) $b^2 = a^2(e^2 - 1)$

Concept: Locus of Point — For an ellipse, the eccentricity $e$ relates the semi-major and semi-minor axes. Since $a < b$, the major axis is along the $y$-axis, so $b$ is the semi-major axis and $a$ is the semi-minor axis.

Step 1: For an ellipse with major axis along $y$, the standard relation is $c^2 = b^2 - a^2$, where $c$ is the focal distance from the centre. …

The key is to never confuse which denominator is larger. In the ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the larger denominator tells you which axis is the major axis. Here $a < b$, so $b$ is larger — that means the major axis is along the $y$-axis (vertical). The eccentricity $e$ is always defined using the semi-major axis $a_{\text{major}}$ and the semi-minor axis $b_{\text{minor}}$, with the relation $b_{\text{minor}}^2 = a_{\text{major}}^2 (1 - e^2)$.

Let’s walk through it.

1. Identify the major and minor axes.

   Since $a < b$, the denominator under $y^2$ is larger, so the ellipse is taller than it is wide. The semi-major axis length is $b$ (along $y$), and the semi-minor axis length is $a$ (along $x$).

2. Recall the standard eccentricity relation for an ellipse.

   For any ellipse, if the semi-major axis is $A$ and the semi-minor axis is $B$, then

where $e$ is the eccentricity ($0 < e < 1$). This comes from the definition $e = \sqrt{1 - \frac{B^2}{A^2}}$.

3. Plug in the correct axes. Here $A = b$ (major) and $B = a$ (minor). So: $a^2 = b^2(1 - e^2)$ …