Given the equation of an ellipse 25x² + 9y² - 150x - 72y + 144 = 0, find the centre, foci and eccentricity.

Steps to Solve

1. Rewrite the equation in standard form

We start with the given equation:

$$ 25x^2 + 9y^2 - 150x - 72y + 144 = 0 $$

Rearranging gives:

$$ 25x^2 - 150x + 9y^2 - 72y + 144 = 0 $$

Now, we will group the $x$ and $y$ terms:

$$ 25(x^2 - 6x) + 9(y^2 - 8y) + 144 = 0 $$

2. Complete the square for the x-terms

To complete the square for $x^2 - 6x$, we take half of $-6$, square it, and add inside and outside the parentheses:

$$ x^2 - 6x \Rightarrow (x - 3)^2 - 9 $$

So we have:

$$ 25[(x - 3)^2 - 9] + 9(y^2 - 8y) + 144 = 0 $$

Which simplifies to:

$$ 25(x - 3)^2 - 225 + 9(y^2 - 8y) + 144 = 0 $$

3. Complete the square for the y-terms

For $y^2 - 8y$, we do the same:

$$ y^2 - 8y \Rightarrow (y - 4)^2 - 16 $$

Substituting gives:

$$ 25(x - 3)^2 - 225 + 9[(y - 4)^2 - 16] + 144 = 0 $$

4. Combine constants and isolate the equation

Now combine constants:

$$ 25(x - 3)^2 + 9(y - 4)^2 - 225 - 144 + 144 = 0 $$

This simplifies to:

$$ 25(x - 3)^2 + 9(y - 4)^2 - 225 = 0 $$

Rearranging yields:

$$ 25(x - 3)^2 + 9(y - 4)^2 = 225 $$

Dividing by 225 gives us the standard form of the ellipse:

$$ \frac{(x - 3)^2}{9} + \frac{(y - 4)^2}{25} = 1 $$

5. Identify the center, foci, and eccentricity

From the standard form:

• The center is at $(h, k) = (3, 4)$.
• The semi-major axis $a = 5$ (since $25 = 5^2$) and semi-minor axis $b = 3$ (since $9 = 3^2$).

The distance of the foci from the center is given by:

$$ c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = \sqrt{16} = 4 $$

Thus, the foci are at:

$$ (h \pm c, k) = (3 \pm 4, 4) = (7, 4) \text{ and } (-1, 4) $$

The eccentricity $e$ is calculated as:

$$ e = \frac{c}{a} = \frac{4}{5} $$