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)}
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
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and More.