Find the points farthest and nearest on the curve `5x^2+5y^2+6xy -8=0` from origin.Also determine the equations of the lines through the origin on which the points are occurring.

Answer

To solve the problem of finding the points farthest and nearest on the curve \(5x^2 + 5y^2 + 6xy - 8 = 0\) from the origin, and determining the equations of the lines through the origin on which these points occur, we will follow these steps:

### Step 1: Rewrite the equation in polar coordinates
We start by substituting \(x = r \cos \theta\) and \(y = r \sin \theta\) into the equation of the curve.

\[ 5(r \cos \theta)^2 + 5(r \sin \theta)^2 + 6(r \cos \theta)(r \sin \theta) - 8 = 0 \]

This simplifies to:

\[ 5r^2 \cos^2 \theta + 5r^2 \sin^2 \theta + 6r^2 \cos \theta \sin \theta - 8 = 0 \]

### Step 2: Factor out \(r^2\)
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\):

\[ 5r^2 + 6r^2 \cos \theta \sin \theta - 8 = 0 \]

### Step 3: Simplify the equation
Now we can rewrite the equation as:

\[ r^2(5 + 3 \sin 2\theta) - 8 = 0 \]

### Step 4: Solve for \(r^2\)
Rearranging gives:

\[ r^2(5 + 3 \sin 2\theta) = 8 \]

Thus,

\[ r^2 = \frac{8}{5 + 3 \sin 2\theta} \]

### Step 5: Find maximum and minimum values of \(r\)
To find the farthest and nearest points, we need to analyze the expression \(5 + 3 \sin 2\theta\). The maximum value of \(\sin 2\theta\) is 1 and the minimum is -1.

- Maximum: \(5 + 3(1) = 8\)
- Minimum: \(5 + 3(-1) = 2\)

### Step 6: Calculate \(r^2\) for maximum and minimum
- For maximum distance:

\[ r^2_{\text{max}} = \frac{8}{8} = 1 \implies r_{\text{max}} = 1 \]

- For minimum distance:

\[ r^2_{\text{min}} = \frac{8}{2} = 4 \implies r_{\text{min}} = 2 \]

### Step 7: Find the corresponding angles
To find the angles corresponding to these distances, we set:

1. For maximum \(r_{\text{max}} = 1\):

\[ 3 \sin 2\theta = 3 \implies \sin 2\theta = 1 \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4} \]

2. For minimum \(r_{\text{min}} = 2\):

\[ 3 \sin 2\theta = -1 \implies \sin 2\theta = -\frac{1}{3} \]

This gives two angles, \(2\theta = \arcsin(-\frac{1}{3})\) and \(2\theta = \pi - \arcsin(-\frac{1}{3})\). Thus, we can find \(\theta\) from these angles.

### Step 8: Calculate the coordinates of the points
1. For \(\theta = \frac{\pi}{4}\):

\[ x = r \cos \theta = 1 \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}, \quad y = r \sin \theta = 1 \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

2. For the other angles, we calculate \(x\) and \(y\) using \(r = 2\).

### Step 9: Determine the equations of the lines
1. For the point \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\):

The slope \(m = \frac{y}{x} = 1\), so the equation is:

\[ y = x \]

2. For the point corresponding to the minimum distance, calculate the slope similarly.

### Summary of Results
- **Nearest Point**: \(\left(2 \cos(\theta), 2 \sin(\theta)\right)\) where \(\theta\) corresponds to \(\arcsin(-\frac{1}{3})\).
- **Farthest Point**: \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\).
- **Equations of Lines**: \(y = x\) for the farthest point and another equation for the nearest point.