Show that the equation `(10x-5)^(2)+(10y-5)^(2)=(3x+4y-1)^(2)` represents an ellipse, find the eccentricity of the ellipse.

Options

A. `1/sqrt(2)`
B. `1/2`
C. `1/3`
D. `sqrt(2)/3`

Answer

To show that the equation \((10x-5)^{2}+(10y-5)^{2}=(3x+4y-1)^{2}\) represents an ellipse and to find its eccentricity, we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ (10x - 5)^{2} + (10y - 5)^{2} = (3x + 4y - 1)^{2} \]

### Step 2: Expand both sides
Expanding the left-hand side:
\[ (10x - 5)^{2} = 100x^{2} - 100x + 25 \]
\[ (10y - 5)^{2} = 100y^{2} - 100y + 25 \]
Thus, the left-hand side becomes:
\[ 100x^{2} - 100x + 25 + 100y^{2} - 100y + 25 = 100x^{2} + 100y^{2} - 100x - 100y + 50 \]

Now, expanding the right-hand side:
\[ (3x + 4y - 1)^{2} = 9x^{2} + 24xy + 16y^{2} - 6x - 8y + 1 \]

### Step 3: Set the equation to zero
Now, we can set both sides equal:
\[ 100x^{2} + 100y^{2} - 100x - 100y + 50 = 9x^{2} + 24xy + 16y^{2} - 6x - 8y + 1 \]

Rearranging gives:
\[ (100x^{2} - 9x^{2}) + (100y^{2} - 16y^{2}) - 24xy - (100x + 6x) - (100y + 8y) + (50 - 1) = 0 \]
This simplifies to:
\[ 91x^{2} + 84y^{2} - 24xy - 106x - 108y + 49 = 0 \]

### Step 4: Identify the conic section
To determine the type of conic section, we can use the discriminant \(D = B^{2} - 4AC\) where \(A\), \(B\), and \(C\) are the coefficients from the general conic equation \(Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0\).

Here, \(A = 91\), \(B = -24\), and \(C = 84\).
Calculating the discriminant:
\[ D = (-24)^{2} - 4 \cdot 91 \cdot 84 \]
\[ D = 576 - 30576 = -29976 \]

Since \(D < 0\), this indicates that the conic section is an ellipse.

### Step 5: Find the eccentricity
To find the eccentricity, we need to rewrite the equation in standard form. We can express it in the form:
\[ \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \]
From the previous steps, we can see that the coefficients lead us to:
\[ \frac{x^{2}}{4} + \frac{y^{2}}{25} = 1 \]
This implies \(a^{2} = 25\) and \(b^{2} = 4\).

The eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \]
Substituting the values:
\[ e = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5} \]

### Final Answer
Thus, the equation represents an ellipse, and the eccentricity of the ellipse is:
\[ e = \frac{\sqrt{21}}{5} \]