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

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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.

\[\Rightarrow {{y}^{2}}=4ax\]

\[x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n};y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\]

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.

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.

Last updated date: 26th Sep 2023

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