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\[PQ\] is a double ordinate of a parabola. Find the locus of its point of trisection.

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Last updated date: 25th Jul 2024
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Answer
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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.
\[\Rightarrow {{y}^{2}}=4ax\]
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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.