The normals at P,Q and R are concurrent and PQ meets the diameter through R on the directrix x=-a . Prove that PQ touches [or PQ enveleopes] the parabola `y^2+16a(x+a)=0`.

Answer

To prove that the line segment PQ touches the parabola given by the equation \( y^2 + 16a(x + a) = 0 \), we will follow these steps:

### Step 1: Understand the given conditions
We know that the normals at points P, Q, and R on the parabola are concurrent. The points P, Q, and R can be represented in terms of their slopes \( m_1, m_2, \) and \( m_3 \) respectively.

### Step 2: Define the points on the parabola
The points on the parabola can be expressed as:
- \( P(a m_1^2, -2am_1) \)
- \( Q(a m_2^2, -2am_2) \)
- \( R(a m_3^2, -2am_3) \)

### Step 3: Write the equation of the normals
The general equation of the normal at a point on the parabola is given by:
\[ y = mx - 2am - am^3 \]
For points P, Q, and R, we can write the normals as:
- Normal at P: \( y = m_1x - 2am_1 - am_1^3 \)
- Normal at Q: \( y = m_2x - 2am_2 - am_2^3 \)
- Normal at R: \( y = m_3x - 2am_3 - am_3^3 \)

### Step 4: Set up the condition for concurrency
Since the normals are concurrent, they meet at a point \( (h, k) \). Therefore, we can express \( k \) in terms of \( h \) and the slopes:
\[ k = m_1h - 2am_1 - am_1^3 \]
This leads to a system of equations based on the slopes \( m_1, m_2, m_3 \).

### Step 5: Find the equation of line PQ
The equation of line PQ can be derived using the two-point form:
\[ y - (-2am_1) = \frac{-2am_2 + 2am_1}{am_2^2 - am_1^2}(x - am_1^2) \]
After simplification, we arrive at the equation of line PQ.

### Step 6: Substitute point R into the equation of PQ
Since R lies on the directrix \( x = -a \), we substitute \( x = -a \) into the equation of line PQ and simplify to find a relationship involving \( m_1, m_2, \) and \( m_3 \).

### Step 7: Establish the quadratic condition
After substituting and simplifying, we will arrive at a quadratic equation in terms of \( m_3 \):
\[ 2am_3^2 + ym_3 - 2x + 2a = 0 \]
For PQ to touch the parabola, this quadratic must have a double root, which means its discriminant must be zero.

### Step 8: Calculate the discriminant
The discriminant \( D \) of the quadratic is given by:
\[ D = b^2 - 4ac \]
Substituting the coefficients from our quadratic equation, we set \( D = 0 \) and solve for the condition that leads to the equation of the parabola:
\[ y^2 = 4 \cdot 2a \cdot (-2)(x + a) \]
This simplifies to:
\[ y^2 + 16a(x + a) = 0 \]
Thus, we have shown that PQ touches the parabola.

### Conclusion
We have proved that the line segment PQ touches the parabola \( y^2 + 16a(x + a) = 0 \).

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