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$PQ$ is a double ordinate of a parabola. Find the locus of its point of trisection.

Last updated date: 26th Mar 2023
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Hint: Use the section formula
$x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n};y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$
In between extremities of double ordinate.
Given that $PQ$ is double ordinate of parabola.
We have to find the locus of points of trisection of double ordinate.
Let’s take the standard equation of parabola.
$\Rightarrow {{y}^{2}}=4ax$

We know that $PQ$ is double ordinate of given parabola.
Therefore, $PQ\bot OA$
We know that any general point on parabola ${{y}^{2}}=4ax$is $\left( x,y \right)=\left( a{{t}^{2}},2at \right)$.
Therefore, $P=\left( a{{t}^{2}},2at \right)$
As $P$ and $Q$ are symmetrical along x-axis.
Therefore, $Q=\left( a{{t}^{2}},-2at \right)$
Let $R$ and $S$ be the point of trisection of double ordinate.
Therefore, $PR=RS=SQ.....\left( i \right)$
Let $R$ divide the $PQ$in the ratio $\dfrac{m}{n}=\dfrac{PR}{RQ}$.
Substitute$RQ=RS+SQ$.
$\dfrac{m}{n}=\dfrac{PR}{RQ}=\dfrac{PS}{RS+SQ}$
Putting $RS=SQ=PR\left[ \text{from equation }\left( i \right) \right]$
$\dfrac{m}{n}=\dfrac{PR}{RQ}=\dfrac{PR}{PR+PR}=\dfrac{PR}{2PR}$
Therefore, we get $\dfrac{m}{n}=\dfrac{1}{2}$
Therefore, $R$ divides $PQ$ in ratio$\dfrac{m}{n}=\dfrac{1}{2}$.
By section formula,
$x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n}....\left( ii \right)$
$y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}....\left( iii \right)$
Here, $P\left( {{x}_{1}},{{y}_{1}} \right)=P\left( a{{t}^{2}},2at \right)$
$Q\left( {{x}_{2}},{{y}_{2}} \right)=Q\left( a{{t}^{2}},-2at \right)$
$R\left( x,y \right)=R\left( h,k \right)$
Putting values in equation$\left( ii \right)$and equation$\left( iii \right)$.
Therefore, $h=\dfrac{1\left( a{{t}^{2}} \right)+2\left( a{{t}^{2}} \right)}{3}$, $k=\dfrac{1\left( -2at \right)+2\left( 2at \right)}{3}$
We get, $h=\dfrac{3a{{t}^{2}}}{3}=a{{t}^{2}}....\left( iv \right)$, $k=\dfrac{2at}{3}....\left( v \right)$
From equation$\left( v \right)$, we get $t=\dfrac{3k}{2a}$
Putting value of $t$ in equation $\left( iv \right)$.
$\Rightarrow h=a{{t}^{2}}$
$h=a{{\left( \dfrac{3k}{2a} \right)}^{2}}$
Therefore we get $9{{k}^{2}}=4ah$
Therefore, locus of point of intersection is $9{{y}^{2}}=4ax$.

Note:
Parabola taken in the problem must be standard parabola. Many students make errors while using section formula and reverse the points.