Find the locus of the point of intersection of the perpendicular tangents of the curve `y^2+4y-6x-2=0` .

Explanation

To find the locus of the point of intersection of the perpendicular tangents of the curve given by the equation \( y^2 + 4y - 6x - 2 = 0 \), we will follow the steps outlined below:

### Step 1: Rewrite the equation of the curve
We start with the equation:
\[ y^2 + 4y - 6x - 2 = 0 \]
We can rearrange this to isolate the terms involving \( x \):
\[ y^2 + 4y = 6x + 2 \]

### Step 2: Complete the square for \( y \)
Next, we complete the square for the \( y \) terms:
\[ y^2 + 4y = (y + 2)^2 - 4 \]
Substituting this back into the equation gives:
\[ (y + 2)^2 - 4 = 6x + 2 \]
Adding 4 to both sides:
\[ (y + 2)^2 = 6x + 6 \]
This simplifies to:
\[ (y + 2)^2 = 6(x + 1) \]

### Step 3: Identify the parabola's standard form
Now, we can recognize that this is in the standard form of a parabola:
\[ (y - k)^2 = 4a(x - h) \]
where \( k = -2 \), \( h = -1 \), and \( 4a = 6 \). Thus, we find:
\[ a = \frac{6}{4} = \frac{3}{2} \]

### Step 4: Find the equation of the directrix
For a parabola of the form \( (y - k)^2 = 4a(x - h) \), the equation of the directrix is given by:
\[ x = h - a \]
Substituting \( h = -1 \) and \( a = \frac{3}{2} \):
\[ x = -1 - \frac{3}{2} = -1 - 1.5 = -2.5 \]

### Step 5: Find the locus of the point of intersection of perpendicular tangents
The locus of the point of intersection of the perpendicular tangents to the parabola is given by the equation of the directrix. Thus, we have:
\[ x = -2.5 \]

### Final Answer
The locus of the point of intersection of the perpendicular tangents of the curve is:
\[ x = -2.5 \]

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