Q.Find the equation for the ellipse that satisfies the given conditions: $b = 3$, $c = 4$, centre at the origin; foci on the $x$-axis.

Concept: Locus of Point — For an ellipse centred at the origin with foci on the $x$-axis, the standard equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$ and $c^2 = a^2 - b^2$.

Steps:

1. Given $b = 3$ and $c = 4$, and foci on the $x$-axis, so $c$ is the distance from centre to each focus.
2. Use the relation $c^2 = a^2 - b^2$: …

The problem gives you $b = 3$, $c = 4$, centre at the origin, and foci on the $x$-axis. That last condition — foci on the $x$-axis — tells you the major axis is horizontal. For an ellipse, the foci always lie on the major axis, so the longer axis is along $x$.

The standard equation for such an ellipse is:

where $a$ is the semi-major axis (half the length of the horizontal axis) and $b$ is the semi-minor axis (half the vertical axis). The foci are at $(\pm c, 0)$, and the relationship between $a$, $b$, and $c$ is:

This comes from the definition: for any point on the ellipse, the sum of distances to the two foci is constant and equals $2a$. The geometry of that definition forces $a > c$ and $a^2 - b^2 = c^2$.

You already know $b = 3$ and $c = 4$. So plug into the relation:

1. Write the relation: $c^2 = a^2 - b^2$.
2. Substitute: $4^2 = a^2 - 3^2$.
3. That gives $16 = a^2 - 9$.
4. So $a^2 = 16 + 9 = 25$.
5. Hence $a = 5$ (the positive root, since it's a length). …