Coordinates of points on curve `5x^(2) - 6xy +5y^(2) - 4 = 0` which are nearest to origin are

Options

A. `((1)/(2),(1)/(2))`
B. `(-(1)/(2),(1)/(2))`
C. `(-(1)/(2),-(1)/(2))`
D. `((1)/(2),-(1)/(2))`

Explanation

To find the coordinates of the points on the curve \(5x^2 - 6xy + 5y^2 - 4 = 0\) that are nearest to the origin, we can follow these steps:

### Step 1: Substitute Polar Coordinates
We start by substituting the polar coordinates into the equation. We let:
\[ x = r \cos \theta \quad \text{and} \quad y = r \sin \theta \]
Substituting these into the given equation:
\[ 5(r \cos \theta)^2 - 6(r \cos \theta)(r \sin \theta) + 5(r \sin \theta)^2 - 4 = 0 \]

### Step 2: Simplify the Equation
Expanding the equation gives:
\[ 5r^2 \cos^2 \theta - 6r^2 \sin \theta \cos \theta + 5r^2 \sin^2 \theta - 4 = 0 \]
We can factor out \(r^2\):
\[ r^2(5 \cos^2 \theta - 6 \sin \theta \cos \theta + 5 \sin^2 \theta) - 4 = 0 \]
This simplifies to:
\[ r^2(5(\cos^2 \theta + \sin^2 \theta) - 6 \sin \theta \cos \theta) - 4 = 0 \]
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ r^2(5 - 6 \sin \theta \cos \theta) - 4 = 0 \]

### Step 3: Rearranging the Equation
Rearranging gives:
\[ r^2 = \frac{4}{5 - 6 \sin \theta \cos \theta} \]
Using the double angle identity \(\sin 2\theta = 2 \sin \theta \cos \theta\):
\[ r^2 = \frac{4}{5 - 3 \sin 2\theta} \]

### Step 4: Minimize \(r^2\)
To minimize \(r\), we need to maximize the denominator \(5 - 3 \sin 2\theta\). The maximum value of \(\sin 2\theta\) is 1, so:
\[ 5 - 3 \cdot 1 = 2 \]
Thus:
\[ r^2_{\text{min}} = \frac{4}{2} = 2 \quad \Rightarrow \quad r_{\text{min}} = \sqrt{2} \]

### Step 5: Finding \(\theta\)
The value of \(\sin 2\theta = 1\) occurs at:
\[ 2\theta = \frac{\pi}{2} + 2k\pi \quad \Rightarrow \quad \theta = \frac{\pi}{4} + k\pi \]
This gives us two angles in the range of \(0\) to \(2\pi\):
\[ \theta = \frac{\pi}{4}, \quad \frac{5\pi}{4} \]

### Step 6: Calculate the Coordinates
Now substituting back to find \(x\) and \(y\):
1. For \(\theta = \frac{\pi}{4}\):
\[ x = r \cos \theta = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \]
\[ y = r \sin \theta = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \]
Coordinates: \((1, 1)\)

2. For \(\theta = \frac{5\pi}{4}\):
\[ x = r \cos \theta = \sqrt{2} \cdot \left(-\frac{1}{\sqrt{2}}\right) = -1 \]
\[ y = r \sin \theta = \sqrt{2} \cdot \left(-\frac{1}{\sqrt{2}}\right) = -1 \]
Coordinates: \((-1, -1)\)

### Final Answer
The coordinates of the points on the curve that are nearest to the origin are:
\[ (1, 1) \quad \text{and} \quad (-1, -1) \]