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:
Show 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}
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 \( (1,2) \):
\[
y - 2 = \frac{5}{4}(x - 1)
\]
\[
\Rightarrow 5x - 4y + 3 = 0
\]
