Let the locus of the midpoint of a chord through the origin of the parabola y^2=4x be the …

Question

Let the locus of the midpoint of a chord through the origin of the parabola $y^2=4x$ be the curve $S$. Let $P$ be any point on $S$. Then the locus of the point which internally divides $OP$ in the ratio $3:1$ is:

Accepted Answer

For a chord of $y^2=4x$ with midpoint $(x_1,y_1)$: $S_1=S_{11}\Rightarrow yy_1-2(x+x_1)=y_1^2-4x_1$. Passing through the origin: $-2x_1=y_1^2-4x_1\Rightarrow y_1^2=2x_1$, so $S: y^2=2x$.

Parametrize $S$ as $P=\left(\dfrac{t^2}{2},t\right)$. The point $Q$ dividing $OP$ in ratio $3:1$:

$$Q=\left(\frac{3t^2}{8},\ \frac{3t}{4}\right) \;\Rightarrow\; x=\frac{3t^2}{8},\ y=\frac{3t}{4}\;\Rightarrow\; t=\frac{4y}{3}.$$

$$x=\frac38\cdot\frac{16y^2}{9}=\frac{2y^2}{3} \;\Rightarrow\; 2y^2=3x.$$

Solutions · 1