Q.Given the ellipse with equation $9x^2 + 25y^2 = 225$, find the eccentricity and foci.

Concept: Ellipse Standard Form — The given equation must be rewritten as $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ to identify $a$ and $b$, then use $e = \sqrt{1 - \frac{b^2}{a^2}}$ (for $a > b$).

Step 1: Convert to standard form

Divide $9x^2 + 25y^2 = 225$ by $225$:

So $a^2 = 25$, $b^2 = 9$, with $a = 5$, $b = 3$. Since $a > b$, the major axis is along the $x$-axis.

Step 2: Find eccentricity

$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ …

The equation $9x^2 + 25y^2 = 225$ is not yet in the standard form of an ellipse. The standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ and $b$ are the semi-major and semi-minor axes. The eccentricity $e$ measures how "stretched" the ellipse is — it's the ratio of the distance from the centre to a focus ($c$) to the semi-major axis ($a$). The foci are the two fixed points inside the ellipse such that the sum of distances from any point on the ellipse to them is constant.

Let's rewrite the given equation.

1. Divide through by 225 to get the standard form:

So $a^2 = 25$ and $b^2 = 9$. Since $25 > 9$, the major axis is along the $x$-axis. Hence $a = 5$ and $b = 3$.

2. Find $c$, the distance from the centre to each focus. For an ellipse, the relationship is $c^2 = a^2 - b^2$ (when $a > b$). This comes from the definition: the foci are at $(\pm c, 0)$, and the sum of distances from a point on the ellipse to the foci is $2a$. Using the point $(a, 0)$ gives $c^2 = a^2 - b^2$.

3. Compute the eccentricity $e$. By definition, $e = \frac{c}{a}$. $e = \frac{4}{5}.$ …