Perpendicular tangents are drawn from an external point P to the parabola `y^2=16(x-3)` Then the locus of point P is

Options

A. `x=1`
B. `x=-1`
C. `x=1/2`
D. `x=2`

Answer

To find the locus of the point P from which perpendicular tangents are drawn to the parabola given by the equation \(y^2 = 16(x - 3)\), we will follow these steps:

### Step 1: Identify the standard form of the parabola
The given parabola \(y^2 = 16(x - 3)\) can be rewritten in the standard form:
\[ y^2 = 4p(x - h) \]
where \(p = 4\) and \(h = 3\). This indicates that the vertex of the parabola is at the point \((3, 0)\).

### Step 2: Determine the focus and directrix of the parabola
For the parabola \(y^2 = 16(x - 3)\):
- The focus is at \((h + p, 0) = (3 + 4, 0) = (7, 0)\).
- The directrix is given by the equation \(x = h - p = 3 - 4 = -1\).

### Step 3: Set up the condition for perpendicular tangents
Let the external point \(P\) have coordinates \((h, k)\). The condition for the tangents from point \(P\) to the parabola to be perpendicular can be derived using the fact that the product of the slopes of the tangents must equal \(-1\).

### Step 4: Use the tangent equation
The equation of the tangent to the parabola \(y^2 = 16(x - 3)\) at a point \((x_1, y_1)\) on the parabola can be given as:
\[ yy_1 = 8(x + 3) \]
If we consider the point \(P(h, k)\), the tangents from \(P\) can be expressed using the general form of the tangent equation.

### Step 5: Find the locus of point P
The condition for perpendicular tangents leads us to the equation:
\[ \frac{(k - 0)^2}{16} + \frac{(h - 3)^2}{4} = 1 \]
This represents an ellipse centered at \((3, 0)\) with semi-major axis \(4\) and semi-minor axis \(2\).

### Step 6: Rearranging the equation
Rearranging the above equation gives us the locus of point \(P\):
\[ \frac{(h - 3)^2}{4} + \frac{k^2}{16} = 1 \]

### Final Result
The locus of point \(P\) is:
\[ \frac{(x - 3)^2}{4} + \frac{y^2}{16} = 1 \]