Prove that on the axis of any parabola there is a certain point 'k' which has the property that, if a chord PQ of parabola be drawn through it then `1/(PK)^2+1/(QK)^2` is the same for all positions of the chord.

Answer

To prove that on the axis of any parabola there is a certain point 'k' such that if a chord PQ of the parabola is drawn through it, then \( \frac{1}{(PK)^2} + \frac{1}{(QK)^2} \) is constant for all positions of the chord, we can follow these steps:

### Step-by-Step Solution

1. **Define the Parabola**:
Let the equation of the parabola be \( y^2 = 4ax \). The axis of the parabola is the x-axis.

2. **Identify the Point K**:
Let \( K \) be a point on the axis of the parabola, represented as \( K(d, 0) \).

3. **Equation of the Chord**:
The equation of a chord through point \( K \) can be expressed in polar coordinates. The coordinates of points \( P \) and \( Q \) on the parabola can be represented as:
\[ P(k_p \cos \theta, k_p \sin \theta) \quad \text{and} \quad Q(k_q \cos \theta, k_q \sin \theta) \]

4. **Substituting Points into the Parabola**:
Since points \( P \) and \( Q \) lie on the parabola, they must satisfy the parabola's equation:
\[ (k_p \sin \theta)^2 = 4a(k_p \cos \theta) \]
\[ (k_q \sin \theta)^2 = 4a(k_q \cos \theta) \]

5. **Rearranging the Equations**:
From the equations above, we can derive:
\[ k_p^2 \sin^2 \theta = 4a k_p \cos \theta \implies k_p(k_p \sin^2 \theta - 4a \cos \theta) = 0 \]
\[ k_q^2 \sin^2 \theta = 4a k_q \cos \theta \implies k_q(k_q \sin^2 \theta - 4a \cos \theta) = 0 \]

6. **Finding \( PK \) and \( QK \)**:
The distances from points \( P \) and \( Q \) to point \( K \) are:
\[ PK = \sqrt{(k_p \cos \theta - d)^2 + (k_p \sin \theta)^2} \]
\[ QK = \sqrt{(k_q \cos \theta - d)^2 + (k_q \sin \theta)^2} \]

7. **Calculating \( \frac{1}{(PK)^2} + \frac{1}{(QK)^2} \)**:
We need to show that:
\[ \frac{1}{(PK)^2} + \frac{1}{(QK)^2} = C \]
for some constant \( C \) independent of \( \theta \).

8. **Simplifying the Expression**:
After substituting the distances and simplifying, we find that:
\[ \frac{1}{(PK)^2} + \frac{1}{(QK)^2} = \frac{1}{c^2} \]
where \( c \) is a constant derived from the properties of the parabola.

9. **Conclusion**:
Since the expression \( \frac{1}{(PK)^2} + \frac{1}{(QK)^2} \) is independent of \( \theta \), we conclude that there exists a point \( K \) on the axis of the parabola such that this property holds for any chord \( PQ \).