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 a 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 locus and chord problems on a parabola: \begin{itemize} \item Use parametric form for simplicity \item Apply section formula carefully \item For chord bisected at a point, use midpoint conditions \end{itemize}

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 \( C \) is: \[ 5x^2 = 8y \]
Step 4: Let the chord of this parabola be bisected at \( (1,2) \). Let the corresponding parameters of the endpoints be \( t_1, t_2 \). Using midpoint formula: \[ \frac{2}{5}(t_1 + t_2) = 1 \Rightarrow t_1 + t_2 = \frac{5}{2} \] \[ \frac{1}{5}(t_1^2 + t_2^2) = 2 \Rightarrow t_1^2 + t_2^2 = 10 \]
Step 5: Slope of the chord: \[ m = \frac{t_1 + t_2}{2} = \frac{5}{4} \] Equation of chord through \( (1,2) \): \[ y - 2 = \frac{5}{4}(x - 1) \] \[ \Rightarrow 5x - 4y + 3 = 0 \]