Locus of Point on Parabola

Given: Point on parabola \( y^2 = 4a x \) is at distance \( 5a \) from focus \( (a, 0) \).

Distance Equation:

\[ (x - a)^2 + y^2 = 25a^2 \Rightarrow (x - a)^2 + 4a x = 25a^2 \Rightarrow x^2 + 2a x - 24a^2 = 0 \]

Solving gives: \( x = 4a \), \( y = 4a \)

✅ Final Answer: \( \boxed{(4a,\ 4a)} \)

Locus of Midpoint of Chords

Given Parabola: \( y^2 = 4x \)

Condition: Chords pass through the vertex \( (0, 0) \)

Let the other end of the chord be \( (x_1, y_1) \), so the midpoint is:

\( M = \left( \frac{x_1}{2}, \frac{y_1}{2} \right) = (h, k) \)

Since the point lies on the parabola: \( y_1^2 = 4x_1 \)

⇒ \( (2k)^2 = 4(2h) \)

⇒ \( 4k^2 = 8h \)

⇒ \( \boxed{k^2 = 2h} \)

✅ Locus of midpoints: \( y^2 = 2x \)