Let O be the vertex of the parabola x^2=4y and Q be any point on it.

Question

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

Accepted Answer

Let $Q=(2t,t^2)$ on $x^2=4y$. $P$ divides $OQ$ in ratio $2:3$:

$$P=\left(\frac{4t}{5},\ \frac{2t^2}{5}\right) \;\Rightarrow\; x=\frac{4t}{5},\ y=\frac{2t^2}{5}.$$

Eliminating $t=\dfrac{5x}{4}$: $y=\dfrac{2}{5}\cdot\dfrac{25x^2}{16}=\dfrac{5x^2}{8}$, i.e. $C: x^2=\dfrac85 y$ (so $4a=\dfrac85,\ 2a=\dfrac45$).

Chord of $x^2=4ay$ bisected at $(x_1,y_1)$: $xx_1-2ay = x_1^2-2ay_1$. With $(x_1,y_1)=(1,2),\ 2a=\dfrac45$:

$$x-\frac45y = 1-\frac85 = -\frac35 \;\Rightarrow\; 5x-4y+3=0.$$

Solutions · 1