Let O be the vertex of the parabola $x^2=4 y$ and Q be any point on it. Let the locus of the point P , which divides the line segment OQ internally in the ratio $2: 3$ be the conic C . Then the equation of the chord of $C$, which is bisected at the point $(1,2)$, is :

Question

Accepted Answer

A. 1. Let $O(0,0)$ be the vertex and $Q$ a point on $x^2=4y$, parametrized as $Q(2t,t^2)$.

2. Let $P(x,y)$ divide $OQ$ internally in the ratio $2:3$. Using the **section formula**:
$$
x=\frac{2(2t)+3(0)}{5}=\frac{4t}{5},\qquad y=\frac{2(t^2)+3(0)}{5}=\frac{2t^2}{5}
$$

3. Eliminate the parameter $t$. From the first equation, $t=\frac{5x}{4}$. Substituting into the second:
$$
y=\frac25\left(\frac{5x}{4}\right)^2=\frac25\cdot\frac{25x^2}{16}=\frac{5x^2}{8}
$$

4. So the **conic C** is:
$$
5x^2=8y
$$

5. The equation of a chord of a conic $S=0$ bisected at a point $(x_1,y_1)$ follows the standard **"T = S1" rule**. Here $S\equiv5x^2-8y$, and $(x_1,y_1)=(1,2)$:
$$
S_1=5(1)^2-8(2)=5-16=-11
$$

6. Construct $T$ by replacing $x^2$ with $xx_1$ and $y$ with $\frac{y+y_1}{2}$:
$$
T=5x(1)-8\left(\frac{y+2}{2}\right)=5x-4(y+2)=5x-4y-8
$$

7. Setting $T=S_1$:
$$
5x-4y-8=-11 \;\Rightarrow\; 5x-4y+3=0
$$

Hence, the answer is **Option A**.

Let O be the vertex of the parabola x2=4yx^2=4 yx2=4y and Q be any point on it. Let the locus of the point P , which divides the line segment OQ internally in the ratio 2:32: 32:3 be the conic C . Then the equation of the chord of CCC, which is bisected at the point (1,2)(1,2)(1,2), is :

Let O be the vertex of the parabola $x^2=4 y$ and Q be any point on it. Let the locus of the point P , which divides the line segment OQ internally in the ratio $2: 3$ be the conic C . Then the equation of the chord of $C$ , which is bisected at the point $(1,2)$ , is :