To solve the problem of finding the farthest and nearest points on the curve \(5x^2 + 5y^2 + 6xy - 8 = 0\) from the origin, we will follow these steps:
### Step 1: Convert the Cartesian equation to Polar coordinates
In polar coordinates, we have:
\[
x = r \cos \theta, \quad y = r \sin \theta
\]
Substituting these into the equation of the curve gives:
\[
5(r \cos \theta)^2 + 5(r \sin \theta)^2 + 6(r \sin \theta)(r \cos \theta) - 8 = 0
\]
This simplifies to:
\[
5r^2 (\cos^2 \theta + \sin^2 \theta) + 6r^2 \sin \theta \cos \theta - 8 = 0
\]
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\), we have:
\[
5r^2 + 6r^2 \sin \theta \cos \theta - 8 = 0
\]
### Step 2: Simplify the equation
We can rewrite the equation as:
\[
r^2 (5 + 6 \sin \theta \cos \theta) = 8
\]
This leads to:
\[
r^2 = \frac{8}{5 + 6 \sin \theta \cos \theta}
\]
### Step 3: Express \(\sin 2\theta\)
Using the double angle identity, we know:
\[
\sin 2\theta = 2 \sin \theta \cos \theta
\]
Thus, we can express \(6 \sin \theta \cos \theta\) as:
\[
3 \sin 2\theta
\]
So the equation becomes:
\[
r^2 = \frac{8}{5 + 3 \sin 2\theta}
\]
### Step 4: Find maximum and minimum values
To find the maximum and minimum values of \(r^2\), we need to analyze the expression \(5 + 3 \sin 2\theta\). The sine function varies between -1 and 1, so:
- The minimum value of \(5 + 3 \sin 2\theta\) occurs when \(\sin 2\theta = -1\):
\[
5 - 3 = 2 \quad \Rightarrow \quad r^2_{\text{max}} = \frac{8}{2} = 4 \quad \Rightarrow \quad r_{\text{max}} = 2
\]
- The maximum value of \(5 + 3 \sin 2\theta\) occurs when \(\sin 2\theta = 1\):
\[
5 + 3 = 8 \quad \Rightarrow \quad r^2_{\text{min}} = \frac{8}{8} = 1 \quad \Rightarrow \quad r_{\text{min}} = 1
\]
### Step 5: Find the angles corresponding to maximum and minimum \(r\)
- For \(r_{\text{max}} = 2\), we have:
\[
3 \sin 2\theta = -3 \quad \Rightarrow \quad \sin 2\theta = -1 \quad \Rightarrow \quad 2\theta = \frac{3\pi}{2} \quad \Rightarrow \quad \theta = \frac{3\pi}{4}
\]
- For \(r_{\text{min}} = 1\), we have:
\[
3 \sin 2\theta = 3 \quad \Rightarrow \quad \sin 2\theta = 1 \quad \Rightarrow \quad 2\theta = \frac{\pi}{2} \quad \Rightarrow \quad \theta = \frac{\pi}{4}
\]
### Step 6: Calculate the points
- For the farthest point:
\[
x = r \cos \theta = 2 \cos \frac{3\pi}{4} = 2 \left(-\frac{1}{\sqrt{2}}\right) = -\sqrt{2}, \quad y = r \sin \theta = 2 \sin \frac{3\pi}{4} = 2 \left(\frac{1}{\sqrt{2}}\right) = \sqrt{2}
\]
Thus, the farthest point is \((- \sqrt{2}, \sqrt{2})\).
- For the nearest point:
\[
x = r \cos \theta = 1 \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad y = r \sin \theta = 1 \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}
\]
Thus, the nearest point is \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\).
### Step 7: Determine the equations of the lines
- For the nearest point \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\):
The line through the origin is:
\[
y = x
\]
- For the farthest point \((- \sqrt{2}, \sqrt{2})\):
The line through the origin is:
\[
y = -x
\]
### Final Answer
- Nearest point: \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) with line \(y = x\)
- Farthest point: \((- \sqrt{2}, \sqrt{2})\) with line \(y = -x\)
