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Let $Q(2t, t^2)$ be a point on the parabola $x^2 = 4y$.
Let $C(h, k)$ be the point which divides $OQ$ in the ratio $2:3$. Then, using the section formula, we have:
$h = \frac{2(2t) + 3(0)}{2+3} = \frac{4t}{5}$ and $k = \frac{2(t^2) + 3(0)}{2+3} = \frac{2t^2}{5}$
From the above equations, we get $t = \frac{5h}{4}$. Substituting this into the equation for $k$, we get:
$k = \frac{2}{5} (\frac{5h}{4})^2 = \frac{2}{5} \cdot \frac{25h^2}{16} = \frac{5h^2}{8}$
Replacing $h$ with $x$ and $k$ with $y$, the locus of $C$ is $5x^2 = 8y$.
The equation of the chord of the curve $5x^2 = 8y$ bisected at the point $(1, 2)$ is given by $T = S_1$, where $T = 5x(1) - 4(y+2)$ and $S_1 = 5(1)^2 - 8(2)$.
So, $5x - 4y - 8 = 5 - 16 = -11$.
Therefore, $5x - 4y - 8 + 11 = 0$, which simplifies to $5x - 4y + 3 = 0$.
Correct Answer: $5x-4y+3=0$
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