The eccentricity of the ellipse $25x^{2}+16y^{2}-150x-175=0$ is:

Question

Accepted Answer

To determine the eccentricity of the given ellipse, we first need to rewrite the equation in the standard form of an ellipse. The given equation is:

$$ 25x^2 + 16y^2 - 150x - 175 = 0. $$

First, we complete the square for the $x$ terms:

1. Group the $x$ terms: $25x^2 - 150x$.
2. Factor out the 25: $25(x^2 - 6x)$.
3. Complete the square: $x^2 - 6x$ becomes $(x - 3)^2 - 9$.
4. Substitute back: $25((x - 3)^2 - 9) = 25(x - 3)^2 - 225$.

Now, substitute this back into the original equation:

$$ 25(x - 3)^2 - 225 + 16y^2 - 175 = 0. $$

Simplify:

$$ 25(x - 3)^2 + 16y^2 = 400. $$

Divide the entire equation by 400 to get it into standard form:

$$ \frac{(x - 3)^2}{16} + \frac{y^2}{25} = 1. $$

This is now in the standard form of an ellipse:

$$ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1, $$

where $a^2 = 16$ and $b^2 = 25$. Here, $b^2 > a^2$, indicating that this is a vertical ellipse.

The formula for the eccentricity $e$ of an ellipse is:

$$ e = \sqrt{1 - \frac{a^2}{b^2}}. $$

Substitute the values:

$$ e = \sqrt{1 - \frac{16}{25}} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6. $$

Thus, the eccentricity $e = 0.6$, which is equivalent to $\frac{3}{5}$.

Therefore, Option 2, $\frac{3}{5}$, is the correct answer.

**Explanation for Incorrect Options:**

1. **Option 1: $ \frac{2}{5} $** - This option is incorrect because the calculation of the eccentricity using the formula $\sqrt{1 - \frac{a^2}{b^2}}$ results in $0.6$, not $0.4$.

2. **Option 3: $ \frac{4}{5} $** - This option is incorrect because it suggests a higher eccentricity than calculated. The correct eccentricity is $0.6$, not $0.8$.

3. **Option 4: $ \frac{1}{5} $** - This option is incorrect because it suggests a lower eccentricity than calculated. The correct eccentricity is $0.6$, not $0.2$.

In conclusion, the correct eccentricity of the ellipse is $\frac{3}{5}$, which is why Option 2 is the correct answer.