If PQ is the focal chord of the parabola `y^(2)=-x and P` is `(-4, 2)`, then the ordinate of the point of intersection of the tangents at P and Q is

Options

A. `(15)/(16)`
B. `(7)/(8)`
C. 4
D. `(17)/(18)`

Answer

To solve the problem, we need to find the ordinate (y-coordinate) of the point of intersection of the tangents at points P and Q on the parabola \( y^2 = -x \), where P is given as \( (-4, 2) \).

### Step-by-Step Solution:

1. **Identify the Parabola and its Focus**:
The equation of the parabola is given as \( y^2 = -x \). This can be rewritten in the standard form \( y^2 = 4px \) where \( p = -\frac{1}{4} \). Thus, the focus of the parabola is at \( (-\frac{1}{4}, 0) \).

**Hint**: Recall that the focus of the parabola \( y^2 = 4px \) is at \( (p, 0) \).

2. **Determine the Parameter for Point P**:
The coordinates of point P are given as \( (-4, 2) \). For a point \( (x, y) \) on the parabola, we can express it in terms of the parameter \( t \) as:
\[ (t^2, 2at) \quad \text{where } a = -\frac{1}{4} \]
Here, we need to find \( t_1 \) such that:
\[ t_1^2 = -4 \quad \text{and} \quad 2at_1 = 2 \]
From \( 2at_1 = 2 \), we have:
\[ t_1 = -4 \]

**Hint**: Use the parametric equations of the parabola to find the parameter corresponding to point P.

3. **Find the Parameter for Point Q**:
Since PQ is a focal chord, the product of the parameters \( t_1 \) and \( t_2 \) is \( -1 \):
\[ t_1 t_2 = -1 \implies (-4)t_2 = -1 \implies t_2 = \frac{1}{4} \]
Now we can find the coordinates of point Q:
\[ Q = \left(t_2^2, 2at_2\right) = \left(\left(\frac{1}{4}\right)^2, 2 \cdot -\frac{1}{4} \cdot \frac{1}{4}\right) = \left(\frac{1}{16}, -\frac{1}{8}\right) \]

**Hint**: Remember that for focal chords, the parameters multiply to -1.

4. **Equation of the Tangent at Point P**:
The slope of the tangent at point P can be found using the derivative of the parabola:
\[ \frac{dy}{dx} = -\frac{1}{2y} \]
At point P \( (x_1, y_1) = (-4, 2) \):
\[ \text{slope} = -\frac{1}{2 \cdot 2} = -\frac{1}{4} \]
The equation of the tangent at P is:
\[ y - 2 = -\frac{1}{4}(x + 4) \implies y = -\frac{1}{4}x + 1 \]

**Hint**: Use the point-slope form of the line to find the tangent equation.

5. **Equation of the Tangent at Point Q**:
Similarly, find the slope at point Q \( (x_2, y_2) = \left(\frac{1}{16}, -\frac{1}{8}\right) \):
\[ \text{slope} = -\frac{1}{2 \cdot -\frac{1}{8}} = 4 \]
The equation of the tangent at Q is:
\[ y + \frac{1}{8} = 4\left(x - \frac{1}{16}\right) \implies y = 4x - \frac{1}{2} - \frac{1}{8} = 4x - \frac{5}{8} \]

**Hint**: Again, use the point-slope form to derive the tangent equation.

6. **Finding the Intersection of the Tangents**:
Now we need to find the intersection of the two tangent lines:
\[ -\frac{1}{4}x + 1 = 4x - \frac{5}{8} \]
Multiplying through by 8 to eliminate fractions:
\[ -2x + 8 = 32x - 5 \implies 34x = 13 \implies x = \frac{13}{34} \]
Substitute \( x \) back into one of the tangent equations to find \( y \):
\[ y = -\frac{1}{4}\left(\frac{13}{34}\right) + 1 = -\frac{13}{136} + 1 = \frac{136 - 13}{136} = \frac{123}{136} \]

**Hint**: Solve the system of equations simultaneously to find the intersection point.

### Final Answer:
The ordinate of the point of intersection of the tangents at P and Q is \( \frac{123}{136} \).