Find the equation of the tangents drawn at the ends of the major axis of the ellipse`9x^(2)+5y^(2)-30y=0`

Explanation

To find the equation of the tangents drawn at the ends of the major axis of the ellipse given by the equation \(9x^2 + 5y^2 - 30y = 0\), we will follow these steps:

### Step 1: Rewrite the equation of the ellipse
We start with the given equation:

\[ 9x^2 + 5y^2 - 30y = 0 \]

We can rearrange this equation by isolating the terms involving \(y\):

\[ 9x^2 + 5(y^2 - 6y) = 0 \]

### Step 2: Complete the square for the \(y\) terms
Next, we complete the square for the expression \(y^2 - 6y\):

\[ y^2 - 6y = (y - 3)^2 - 9 \]

Substituting this back into the equation gives:

\[ 9x^2 + 5((y - 3)^2 - 9) = 0 \]

This simplifies to:

\[ 9x^2 + 5(y - 3)^2 - 45 = 0 \]

### Step 3: Rearranging the equation
Now, we can rearrange this to isolate the constant:

\[ 9x^2 + 5(y - 3)^2 = 45 \]

### Step 4: Divide by 45 to get the standard form
Dividing the entire equation by 45, we have:

\[ \frac{x^2}{5} + \frac{(y - 3)^2}{9} = 1 \]

### Step 5: Identify the parameters of the ellipse
From the standard form of the ellipse \(\frac{x^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we can identify:

- \(a^2 = 5\) (thus \(a = \sqrt{5}\))
- \(b^2 = 9\) (thus \(b = 3\))
- The center of the ellipse is at \((0, 3)\).

### Step 6: Determine the ends of the major axis
Since \(b > a\), the major axis is vertical. The ends of the major axis are at:

\[ (0, 3 + 3) = (0, 6) \quad \text{and} \quad (0, 3 - 3) = (0, 0) \]

### Step 7: Find the equations of the tangents
The tangents at the ends of the major axis (which are vertical lines) can be determined as follows:

1. For the point \((0, 6)\), the tangent line is:
\[ y = 6 \]

2. For the point \((0, 0)\), the tangent line is:
\[ y = 0 \]

### Final Answer
Thus, the equations of the tangents drawn at the ends of the major axis of the ellipse are:

\[ y = 6 \quad \text{and} \quad y = 0 \]

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