Find the point on the axis of the parabola `3y^2+4y-6x+8 =0` from where three distinct normals can be drawn.

Answer

To find the point on the axis of the parabola \(3y^2 + 4y - 6x + 8 = 0\) from where three distinct normals can be drawn, we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the given equation:
\[ 3y^2 + 4y - 6x + 8 = 0 \]
Rearranging gives:
\[ 3y^2 + 4y = 6x - 8 \]
Now, we factor out the 3 from the left side:
\[ 3(y^2 + \frac{4}{3}y) = 6x - 8 \]
Next, we complete the square for the \(y\) terms. The coefficient of \(y\) is \(\frac{4}{3}\). Half of this is \(\frac{2}{3}\), and squaring it gives \(\frac{4}{9}\). We add and subtract this inside the bracket:
\[ 3\left(y^2 + \frac{4}{3}y + \frac{4}{9} - \frac{4}{9}\right) = 6x - 8 \]
This simplifies to:
\[ 3\left((y + \frac{2}{3})^2 - \frac{4}{9}\right) = 6x - 8 \]
Distributing the 3 gives:
\[ 3(y + \frac{2}{3})^2 - \frac{4}{3} = 6x - 8 \]
Now, adding \(\frac{4}{3}\) to both sides:
\[ 3(y + \frac{2}{3})^2 = 6x - 8 + \frac{4}{3} \]
Converting \(-8\) to a fraction with a denominator of 3 gives \(-\frac{24}{3}\):
\[ 3(y + \frac{2}{3})^2 = 6x - \frac{24}{3} + \frac{4}{3} \]
This simplifies to:
\[ 3(y + \frac{2}{3})^2 = 6x - \frac{20}{3} \]
Dividing everything by 3:
\[ (y + \frac{2}{3})^2 = 2x - \frac{20}{9} \]
This is now in the standard form of a parabola.

### Step 2: Identify the vertex and axis of the parabola
From the standard form \((y - k)^2 = 4p(x - h)\), we can identify:
- Vertex \((h, k) = \left(\frac{10}{9}, -\frac{2}{3}\right)\)
- The axis of the parabola is given by \(y = -\frac{2}{3}\).

### Step 3: Determine the conditions for three distinct normals
For three distinct normals to be drawn from a point on the axis, the condition is:
\[ h > 2a \]
where \(a\) is the distance from the vertex to the focus. From the standard form, we have \(4p = 2\), thus \(p = \frac{1}{2}\) and \(a = \frac{1}{2}\).

### Step 4: Substitute and solve the inequality
Substituting \(a\) into the inequality:
\[ h > 2 \times \frac{1}{2} \implies h > 1 \]
The x-coordinate \(h\) of any point on the axis is given by:
\[ h = x - \frac{10}{9} \]
Thus:
\[ x - \frac{10}{9} > 1 \implies x > 1 + \frac{10}{9} = \frac{19}{9} \]

### Step 5: Identify the point on the axis
The y-coordinate of any point on the axis is fixed at:
\[ y = -\frac{2}{3} \]
Thus, the point on the axis from where three distinct normals can be drawn is:
\[ \left(h, -\frac{2}{3}\right) \text{ where } h > \frac{19}{9} \]
So, we can choose any \(h\) greater than \(\frac{19}{9}\), for example, \(h = \frac{20}{9}\).

### Final Answer
The point on the axis of the parabola from where three distinct normals can be drawn is:
\[ \left(\frac{20}{9}, -\frac{2}{3}\right) \]