Q.Find the equation for the ellipse that satisfies the given conditions: Length of minor axis $16$, foci $(0, \pm 6)$.

Concept: Locus of Point (Ellipse)

The foci are on the $y$-axis, so the ellipse is vertical:

Step 1 – Identify given values

Length of minor axis $= 2b = 16 \Rightarrow b = 8$.

Foci: $(0, \pm c) = (0, \pm 6) \Rightarrow c = 6$.

Step 2 – Relate $a$, $b$, $c$

For an ellipse: $c^2 = a^2 - b^2$. …

The key here is to first figure out which axis the ellipse is stretched along. The foci are at $(0, \pm 6)$ — both on the y-axis. That tells you the major axis is vertical. In an ellipse, the foci always lie on the major axis, so the longer axis runs along the y-direction.

The minor axis is given as length 16. That means the distance across the ellipse along the shorter (horizontal) direction is 16. So the semi-minor axis $b$ is half of that: $b = 8$.

Now, for any ellipse, the relationship between the semi-major axis $a$, the semi-minor axis $b$, and the distance from the center to each focus $c$ is:

when the major axis is vertical. (If the major axis were horizontal, it would be $c^2 = a^2 - b^2$ as well — the formula is the same, but you need to be careful which is $a$ and which is $b$.)

Here, each focus is 6 units from the center (the origin), so $c = 6$. We already have $b = 8$. Plug into the relation:

So $a = 10$ (the positive root, since it's a length). …