If `P S Q` is a focal chord of the parabola `y^2=8x` such that `S P=6` , then the length of `S Q` is 6 (b) 4 (c) 3 (d) none of these

Options

A. 6
B. 4
C. 3
D. none of these

Explanation

To solve the problem, we need to find the length of \( S Q \) given that \( P S Q \) is a focal chord of the parabola \( y^2 = 8x \) and \( S P = 6 \).

### Step-by-step Solution:

1. **Identify the parameters of the parabola**:
The given parabola is \( y^2 = 8x \). This can be rewritten in the standard form \( y^2 = 4ax \), where \( 4a = 8 \). Thus, we find \( a = 2 \).

2. **Use the property of focal chords**:
For a focal chord \( P S Q \) of a parabola, the lengths \( S P \), \( S Q \), and \( 2a \) are related in such a way that:
\[ \frac{1}{S P} + \frac{1}{S Q} = \frac{2}{2a} \]
Here, \( 2a = 4 \) since \( a = 2 \).

3. **Substitute the known values**:
We know \( S P = 6 \) and \( 2a = 4 \). Substituting these values into the equation gives:
\[ \frac{1}{6} + \frac{1}{S Q} = \frac{2}{4} \]
Simplifying \( \frac{2}{4} \) gives \( \frac{1}{2} \).

4. **Set up the equation**:
Now we have:
\[ \frac{1}{6} + \frac{1}{S Q} = \frac{1}{2} \]

5. **Solve for \( \frac{1}{S Q} \)**:
Rearranging the equation, we get:
\[ \frac{1}{S Q} = \frac{1}{2} - \frac{1}{6} \]
To subtract these fractions, we need a common denominator. The least common multiple of 2 and 6 is 6:
\[ \frac{1}{2} = \frac{3}{6} \]
Therefore:
\[ \frac{1}{S Q} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]

6. **Find \( S Q \)**:
Taking the reciprocal gives:
\[ S Q = 3 \]

### Final Answer:
The length of \( S Q \) is \( 3 \).