The locus of the mid-point of all chords of the parabola \( y^2 = 4x \) which are drawn through its vertex is:
Show Hint
- \( y^2 = 8x \)
- \( y^2 = 2x \)
- \( y^2 + 4y^2 = 16 \)
- \( x^2 = 2y \)
The Correct Option is B
Solution and Explanation
Step 1: Equation of the parabola.
The equation of the given parabola is \( y^2 = 4x \). This is a standard form of the parabola with its vertex at the origin and axis along the x-axis.
Step 2: Chord passing through the vertex.
Consider any chord passing through the vertex. Let the coordinates of the endpoints of the chord be \( P(x_1, y_1) \) and \( Q(x_2, y_2) \). The midpoint \( M \) of this chord is given by the average of the coordinates of \( P \) and \( Q \), i.e.,
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right).
\]
Step 3: Equation of the midpoint.
For the midpoint of the chord, the equation of the locus is derived using the properties of the parabola. The midpoints of all such chords lie on the parabola \( y^2 = 2x \).
Step 4: Conclusion.
Thus, the locus of the mid-point of all such chords is given by the equation \( y^2 = 2x \), and the correct answer is (b).
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