Question

If \( O \) is the vertex of the parabola \( x^2 = 4y \), \( Q \) is a point on the parabola. If \( C \) is the locus of the point which divides \( OQ \) in the ratio \( 2:3 \), then the equation of the chord of \( C \) which is bisected at the point \( (1,2) \) is:

• A. \(5x + 4y + 3 = 0\)
• B. \(5x - 4y - 3 = 0\)
• C. \(5x - 4y + 3 = 0\)
• D. \(5x + 4y - 3 = 0\)

Hint:

For parabola locus problems:
Use parametric coordinates for simplicity
Apply section formula carefully
Chord bisected at a point uses midpoint conditions

Answer 1

The Correct Option is C

Step 1: Parametric coordinates of a point \( Q \) on the parabola \( x^2 = 4y \) are: \[ Q(2t, t^2) \] Vertex \( O = (0,0) \).
Step 2: Point \( C \) divides \( OQ \) internally in the ratio \(2:3\): \[ C\left(\frac{2}{5}\cdot 2t, \frac{2}{5}\cdot t^2\right) = \left(\frac{4t}{5}, \frac{2t^2}{5}\right) \]
Step 3: Eliminate \( t \) to find the locus of \( C \): \[ t = \frac{5x}{4} \] \[ y = \frac{2}{5}\left(\frac{25x^2}{16}\right) = \frac{5x^2}{8} \] Hence, the locus of \( C \) is: \[ 5x^2 = 8y \]
Step 4: Let the chord of this parabola be bisected at \( (1,2) \). For a parabola \( 5x^2 = 8y \), the slope of the chord bisected at a point is: \[ m = \frac{t_1 + t_2}{2} \] Using midpoint condition, the slope is: \[ m = \frac{5}{4} \]
Step 5: Equation of the chord passing through \( (1,2) \): \[ y - 2 = \frac{5}{4}(x - 1) \] \[ \Rightarrow 5x - 4y + 3 = 0 \]