$P Q$ is a chord of the parabola which subtends a right angle at the vertex. Then locus of the centroid of triangle $P S Q$ , where $S$ is the focus of a given parabola, is

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Solution

Step 1. Determine the slopes of $O P$ and $O Q$ .
It is given that $P Q$ is a chord of the parabola that subtends a right angle at the vertex.
Assume the parabola is of the form $x^{2} = 4 y$ and $S (0, 1)$ .
Consider the two points $P$ and $Q$ has coordinates $(2 t_{1}, {t_{1}}^{2})$ and $(2 t_{2}, {t_{2}}^{2})$ .
Then, the slope of $O P$ is as follows:
$\begin{array}{rcl} Slope of O P & = & \frac{{t_{1}}^{2}}{2 t_{1}} \\ \Rightarrow & Slope of O P = \frac{t_{1}}{2} \end{array}$
The slope of $O Q$ is as follows:
$\begin{array}{rcl} Slope of O Q & = & \frac{{t_{2}}^{2}}{2 t_{2}} \\ \Rightarrow & Slope of O Q = \frac{t_{2}}{2} \end{array}$
Hence, the slope of $O P$ is $\frac{t_{1}}{2}$ and $O Q$ is $\frac{t_{2}}{2}$ .
Step 2. Determine the value of $h$ and $k$ in terms of $t_{1}$ and $t_{2}$ .
It is given that $O P$ is perpendicular to $O Q$ with $O$ as the origin.
Since $O P$ is perpendicular to $O Q$ , $\begin{array}{rcl} Slope of O P \times Slope of O Q & = & - 1 \end{array}$ .
$\begin{array}{rcl} \frac{t_{1}}{2} \times \frac{t_{2}}{2} & = & - 1 \\ \Rightarrow & \frac{t_{1} t_{2}}{4} = - 1 \\ \Rightarrow & t_{1} t_{2} = - 4 \end{array}$
Consider the centroid of the parabola to be denoted by $C (h, k)$ and it is given by $C (h, k) = (\frac{2 t_{1} + 2 t_{2} + 0}{3}, \frac{{t_{1}}^{2} + {t_{2}}^{2} + 1}{3})$ .
The value of $h$ and $k$ is as follows:
$\begin{array}{rcl} h & = & \frac{2 t_{1} + 2 t_{2} + 0}{3} \\ \Rightarrow & \frac{3 h}{2} = t_{1} + t_{2} \dots (1) \\ k & = & \frac{{t_{1}}^{2} + {t_{2}}^{2} + 1}{3} \\ \Rightarrow & 3 k - 1 = {t_{1}}^{2} + {t_{2}}^{2} \dots (2) \end{array}$
Hence, the value of $h$ in terms of $t_{1}$ and $t_{2}$ is is $\begin{array}{rcl} \frac{3 h}{2} & = & t_{1} + t_{2} \end{array}$ and $k$ in terms of $t_{1}$ and $t_{2}$ is $\begin{array}{rcl} 3 k - 1 & = & {t_{1}}^{2} + {t_{2}}^{2} \end{array}$ .
Step 3. Determine the locus of the centroid.
Take square on both sides of the equation $(1)$ .
$\begin{array}{rcl} \frac{9 h^{2}}{4} & = & {(t_{1} + t_{2})}^{2} \\ \Rightarrow & \frac{9 h^{2}}{4} = {t_{1}}^{2} + {t_{2}}^{2} + 2 t_{1} t_{2} \\ \Rightarrow & \frac{9 h^{2}}{4} = {t_{1}}^{2} + {t_{2}}^{2} + 2 (- 4) [∴ t_{1} t_{2} = - 4] \\ \Rightarrow & \frac{9 h^{2}}{4} + 8 = {t_{1}}^{2} + {t_{2}}^{2} \dots (3) \end{array}$
Equate equation $(2)$ and $(3)$ .
$\begin{array}{rcl} 3 k - 1 & = & \frac{9 h^{2}}{4} + 8 \\ \Rightarrow & 12 k - 4 = 9 h^{2} + 32 \\ \Rightarrow & 9 h^{2} = 12 k - 36 \\ \Rightarrow & 3 h^{2} = 4 k - 12 \\ \Rightarrow & h^{2} = \frac{4}{3} (k - 3) \end{array}$
Hence, the locus of the centroid is $x^{2} = \frac{4}{3} (y - 3)$ .

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