If a point P on `y^2x`, the foot of the perpendicular from P on the directrix and the focus form an equilateral traingle , then the coordinates of P may be

Options

A. `(3,-2sqrt3)`
B. `(-3,2sqrt3)`
C. `(3,2sqrt3)`
D. `(-3,-2sqrt3)`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Identify the Parabola and its Properties
The given parabola is \( y^2 = 4x \). From the standard form of a parabola \( y^2 = 4ax \), we can identify that \( a = 1 \).

**Hint:** Recall that for the parabola \( y^2 = 4ax \), the focus is at \( (a, 0) \) and the directrix is the line \( x = -a \).

### Step 2: Determine the Focus and Directrix
For the parabola \( y^2 = 4x \):
- The focus is at \( (1, 0) \).
- The directrix is the line \( x = -1 \).

**Hint:** The focus is always located at \( (a, 0) \) and the directrix is at \( x = -a \).

### Step 3: Parametric Representation of Points on the Parabola
Any point \( P \) on the parabola can be expressed in parametric form as:
\[ P(t) = (at^2, 2at) = (t^2, 2t) \]
Here, since \( a = 1 \), we have:
\[ P(t) = (t^2, 2t) \]

**Hint:** Use the parametric form to express points on the parabola, where \( t \) can take any real value.

### Step 4: Find the Foot of the Perpendicular from Point P to the Directrix
The foot of the perpendicular from point \( P(t) \) to the directrix \( x = -1 \) will have the same y-coordinate as \( P(t) \) and an x-coordinate of -1. Thus, the foot of the perpendicular, denoted as \( B \), is:
\[ B = (-1, 2t) \]

**Hint:** The foot of the perpendicular to a vertical line will have the same y-coordinate as the point from which the perpendicular is drawn.

### Step 5: Form the Equilateral Triangle
We need to form an equilateral triangle with vertices at \( P(t) \), \( B \), and the focus \( A(1, 0) \). For an equilateral triangle, the angles between the lines connecting these points must be \( 60^\circ \).

### Step 6: Calculate the Slopes
1. **Slope of line \( AB \)** (from \( A(1, 0) \) to \( B(-1, 2t) \)):
\[ \text{slope of } AB = \frac{2t - 0}{-1 - 1} = \frac{2t}{-2} = -t \]

2. **Slope of line \( AP \)** (from \( A(1, 0) \) to \( P(t) \)):
\[ \text{slope of } AP = \frac{2t - 0}{t^2 - 1} = \frac{2t}{t^2 - 1} \]

### Step 7: Use the Tangent of the Angle Formula
For the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \):
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Here, \( m_1 = -t \) and \( m_2 = \frac{2t}{t^2 - 1} \). Since \( \theta = 60^\circ \), we have \( \tan 60^\circ = \sqrt{3} \).

Setting up the equation:
\[ \sqrt{3} = \left| \frac{-t - \frac{2t}{t^2 - 1}}{1 + \left(-t\right) \left(\frac{2t}{t^2 - 1}\right)} \right| \]

**Hint:** Remember to simplify the expression carefully and solve for \( t \).

### Step 8: Solve for \( t \)
After simplifying the equation, you will find that \( t = \pm \sqrt{3} \).

**Hint:** You may need to cross-multiply and rearrange terms to isolate \( t \).

### Step 9: Find the Coordinates of Point P
Substituting \( t = \sqrt{3} \) and \( t = -\sqrt{3} \) back into the parametric equations for \( P(t) \):
1. For \( t = \sqrt{3} \):
\[ P(\sqrt{3}) = \left((\sqrt{3})^2, 2\sqrt{3}\right) = (3, 2\sqrt{3}) \]
2. For \( t = -\sqrt{3} \):
\[ P(-\sqrt{3}) = \left((- \sqrt{3})^2, 2(-\sqrt{3})\right) = (3, -2\sqrt{3}) \]

### Final Answer
The coordinates of point \( P \) may be:
\[ (3, 2\sqrt{3}) \text{ and } (3, -2\sqrt{3}) \]

**Hint:** Always check if both points satisfy the condition of forming an equilateral triangle with the given focus and directrix.