# Find the locus of the midpoint of the chord of the parabola \[{{y}^{2}}=4ax\], which passes through the point \[\left( 3b,b \right)\].

Answer

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Hint: Write the equation of chord and satisfy the given point and use formula for midpoint which is \[x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\].

Complete step-by-step answer:

We are given a chord of the parabola which passes through the point \[\left( 3b,b \right)\].

Here, we have to find the locus of midpoint of a given chord.

Let the midpoint of the given chord be \[\left( h,k \right)\].

We know that any general point on parabola \[P\left( t \right)\] is \[\left( a{{t}^{2}},2at \right)\].

So, we get point \[R\left( {{t}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)\] and point \[Q\left( {{t}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)\]

We know that equation of any line passing through \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is

\[\left( y-{{y}_{1}} \right)=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}\left( x-{{x}_{1}} \right)\]

Therefore, we get equation of chord passing through \[R\left( {{t}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)\] and point \[Q\left( {{t}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)\] as,

\[\left( y-2a{{t}_{1}} \right)=\dfrac{\left( 2a{{t}_{2}}-2a{{t}_{1}} \right)}{\left( at_{2}^{2}-at_{1}^{2} \right)}\left( x-at_{1}^{2} \right)\]

By cancelling the like terms, we get,

\[\left( y-2a{{t}_{1}} \right)=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( t_{2}^{2}-t_{1}^{2} \right)}\left( x-at_{1}^{2} \right)\]

Since we know that,

\[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)\]

Therefore, we get,

\[\left( y-2a{{t}_{1}} \right)=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( {{t}_{2}}-{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right)}\left( x-at_{1}^{2} \right)\]

By cancelling the like terms, we get,

\[\left( y-2a{{t}_{1}} \right)=\dfrac{2\left( x-at_{1}^{2} \right)}{\left( {{t}_{2}}+{{t}_{1}} \right)}\]

After cross multiplying the above equation, we get,

\[\left( y-2a{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right)=2\left( x-at_{1}^{2} \right)\]

Simplifying the equation, we get,

\[y\left( {{t}_{2}}+{{t}_{1}} \right)-2a{{t}_{1}}{{t}_{2}}-2at_{1}^{2}=2x-2at_{1}^{2}\]

Finally, we get

\[y\left( {{t}_{2}}+{{t}_{1}} \right)-2a{{t}_{1}}{{t}_{2}}=2x.....\left( i \right)\]

Now, we know that midpoint say \[\left( x,y \right)\] of any line joining points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\]is:

\[x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}\] and \[y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\]

Therefore, we get the midpoint \[\left( h,k \right)\] of chord joining \[R\left( {{t}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)\] and point \[Q\left( {{t}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)\] as:

\[h=\dfrac{at_{1}^{2}+at_{2}^{2}}{2}....\left( ii \right)\]

\[k=\dfrac{2a{{t}_{1}}+2a{{t}_{2}}}{2}....\left( iii \right)\]

Taking 2a common from equation \[\left( iii \right)\], we get,

\[k=\dfrac{2a\left( {{t}_{1}}+{{t}_{2}} \right)}{2}\].

Therefore, we get,

\[\dfrac{k}{a}=\left( {{t}_{1}}+{{t}_{2}} \right)....\left( iv \right)\]

Taking ‘a’ common from equation\[\left( ii \right)\], we get,

\[h=\dfrac{a\left( t_{1}^{2}+t_{2}^{2} \right)}{2}\]

Or, \[\dfrac{2h}{a}=t_{1}^{2}+t_{2}^{2}\]

Since, we know that \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}\]

Now, we subtract 2ab from both sides. We get,

\[{{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab\]

Therefore, we get,

\[\dfrac{2h}{a}={{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}-2{{t}_{1}}{{t}_{2}}\]

Now, we put the value of \[\left( {{t}_{1}}+{{t}_{2}} \right)\] from equation (iv). We get,

\[\dfrac{2h}{a}={{\left( \dfrac{k}{a} \right)}^{2}}-2{{t}_{1}}{{t}_{2}}\]

Or \[2{{t}_{1}}{{t}_{2}}=\dfrac{{{k}^{2}}}{{{a}^{2}}}-\dfrac{2h}{a}\]

By dividing both sides by 2, we get,

\[{{t}_{1}}{{t}_{2}}=\dfrac{{{k}^{2}}}{2{{a}^{2}}}-\dfrac{h}{a}.....\left( v \right)\]

Now, we will put the values of \[\left( {{t}_{1}}+{{t}_{2}} \right)\] and \[\left( {{t}_{1}}{{t}_{2}} \right)\] from equation \[\left( iv \right)\]and \[\left( v \right)\] in equation \[\left( i \right)\].

We get,

\[y\left( {{t}_{2}}+{{t}_{1}} \right)-2a\left( {{t}_{1}}{{t}_{2}} \right)=2x\]

\[\Rightarrow y\left( \dfrac{k}{a} \right)-2a\left( \dfrac{{{k}^{2}}}{2{{a}^{2}}}-\dfrac{h}{a} \right)=2x\]

\[\Rightarrow \dfrac{yk}{a}-2a\left( \dfrac{{{k}^{2}}-2ha}{2{{a}^{2}}} \right)=2x\]

By cancelling the like terms, we get,

\[\Rightarrow \dfrac{yk}{a}-\dfrac{\left( {{k}^{2}}-2ha \right)}{a}=2x\]

After simplifying and cross-multiplying above equation, we get,

\[\left( yk \right)-\left( {{k}^{2}}-2ha \right)=2xa\]

Now, we are given that this chord passes through point \[\left( 3b,b \right)\].

Therefore, we will put \[x=3b\] and \[y=b\].

We get,

\[bk-\left( {{k}^{2}}-2ha \right)=2\left( 3b \right)a\]

\[\Rightarrow bk-{{k}^{2}}+2ha=6ab\]

By transposing all the terms to one side,

We get,

\[{{k}^{2}}-bk-2ha+6ab=0\]

Now, to get the locus, we will replace h by x and k by y. We get,

\[{{y}^{2}}-by-2ax+6ab=0\]

So, the locus of the midpoint of chord passing through \[\left( 3b,b \right)\] is \[{{y}^{2}}-by-2ax+6ab=0\].

Note: In these types of questions, we can directly write the equation of chord with respect to mid-point say \[\left( {{x}_{1}},{{y}_{1}} \right)\] which is \[\left( y-{{y}_{1}} \right)=\dfrac{2a}{{{y}_{1}}}\left( x-{{x}_{1}} \right)\] and put \[\left( h,k \right)\] in place of \[\left( {{x}_{1}},{{y}_{1}} \right)\] and point through which chord in passing [here (3b,b)] in place of \[\left( x,y \right)\] to get locus of midpoint of chord.

Complete step-by-step answer:

We are given a chord of the parabola which passes through the point \[\left( 3b,b \right)\].

Here, we have to find the locus of midpoint of a given chord.

Let the midpoint of the given chord be \[\left( h,k \right)\].

We know that any general point on parabola \[P\left( t \right)\] is \[\left( a{{t}^{2}},2at \right)\].

So, we get point \[R\left( {{t}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)\] and point \[Q\left( {{t}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)\]

We know that equation of any line passing through \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is

\[\left( y-{{y}_{1}} \right)=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}\left( x-{{x}_{1}} \right)\]

Therefore, we get equation of chord passing through \[R\left( {{t}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)\] and point \[Q\left( {{t}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)\] as,

\[\left( y-2a{{t}_{1}} \right)=\dfrac{\left( 2a{{t}_{2}}-2a{{t}_{1}} \right)}{\left( at_{2}^{2}-at_{1}^{2} \right)}\left( x-at_{1}^{2} \right)\]

By cancelling the like terms, we get,

\[\left( y-2a{{t}_{1}} \right)=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( t_{2}^{2}-t_{1}^{2} \right)}\left( x-at_{1}^{2} \right)\]

Since we know that,

\[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)\]

Therefore, we get,

\[\left( y-2a{{t}_{1}} \right)=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( {{t}_{2}}-{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right)}\left( x-at_{1}^{2} \right)\]

By cancelling the like terms, we get,

\[\left( y-2a{{t}_{1}} \right)=\dfrac{2\left( x-at_{1}^{2} \right)}{\left( {{t}_{2}}+{{t}_{1}} \right)}\]

After cross multiplying the above equation, we get,

\[\left( y-2a{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right)=2\left( x-at_{1}^{2} \right)\]

Simplifying the equation, we get,

\[y\left( {{t}_{2}}+{{t}_{1}} \right)-2a{{t}_{1}}{{t}_{2}}-2at_{1}^{2}=2x-2at_{1}^{2}\]

Finally, we get

\[y\left( {{t}_{2}}+{{t}_{1}} \right)-2a{{t}_{1}}{{t}_{2}}=2x.....\left( i \right)\]

Now, we know that midpoint say \[\left( x,y \right)\] of any line joining points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\]is:

\[x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}\] and \[y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\]

Therefore, we get the midpoint \[\left( h,k \right)\] of chord joining \[R\left( {{t}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)\] and point \[Q\left( {{t}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)\] as:

\[h=\dfrac{at_{1}^{2}+at_{2}^{2}}{2}....\left( ii \right)\]

\[k=\dfrac{2a{{t}_{1}}+2a{{t}_{2}}}{2}....\left( iii \right)\]

Taking 2a common from equation \[\left( iii \right)\], we get,

\[k=\dfrac{2a\left( {{t}_{1}}+{{t}_{2}} \right)}{2}\].

Therefore, we get,

\[\dfrac{k}{a}=\left( {{t}_{1}}+{{t}_{2}} \right)....\left( iv \right)\]

Taking ‘a’ common from equation\[\left( ii \right)\], we get,

\[h=\dfrac{a\left( t_{1}^{2}+t_{2}^{2} \right)}{2}\]

Or, \[\dfrac{2h}{a}=t_{1}^{2}+t_{2}^{2}\]

Since, we know that \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}\]

Now, we subtract 2ab from both sides. We get,

\[{{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab\]

Therefore, we get,

\[\dfrac{2h}{a}={{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}-2{{t}_{1}}{{t}_{2}}\]

Now, we put the value of \[\left( {{t}_{1}}+{{t}_{2}} \right)\] from equation (iv). We get,

\[\dfrac{2h}{a}={{\left( \dfrac{k}{a} \right)}^{2}}-2{{t}_{1}}{{t}_{2}}\]

Or \[2{{t}_{1}}{{t}_{2}}=\dfrac{{{k}^{2}}}{{{a}^{2}}}-\dfrac{2h}{a}\]

By dividing both sides by 2, we get,

\[{{t}_{1}}{{t}_{2}}=\dfrac{{{k}^{2}}}{2{{a}^{2}}}-\dfrac{h}{a}.....\left( v \right)\]

Now, we will put the values of \[\left( {{t}_{1}}+{{t}_{2}} \right)\] and \[\left( {{t}_{1}}{{t}_{2}} \right)\] from equation \[\left( iv \right)\]and \[\left( v \right)\] in equation \[\left( i \right)\].

We get,

\[y\left( {{t}_{2}}+{{t}_{1}} \right)-2a\left( {{t}_{1}}{{t}_{2}} \right)=2x\]

\[\Rightarrow y\left( \dfrac{k}{a} \right)-2a\left( \dfrac{{{k}^{2}}}{2{{a}^{2}}}-\dfrac{h}{a} \right)=2x\]

\[\Rightarrow \dfrac{yk}{a}-2a\left( \dfrac{{{k}^{2}}-2ha}{2{{a}^{2}}} \right)=2x\]

By cancelling the like terms, we get,

\[\Rightarrow \dfrac{yk}{a}-\dfrac{\left( {{k}^{2}}-2ha \right)}{a}=2x\]

After simplifying and cross-multiplying above equation, we get,

\[\left( yk \right)-\left( {{k}^{2}}-2ha \right)=2xa\]

Now, we are given that this chord passes through point \[\left( 3b,b \right)\].

Therefore, we will put \[x=3b\] and \[y=b\].

We get,

\[bk-\left( {{k}^{2}}-2ha \right)=2\left( 3b \right)a\]

\[\Rightarrow bk-{{k}^{2}}+2ha=6ab\]

By transposing all the terms to one side,

We get,

\[{{k}^{2}}-bk-2ha+6ab=0\]

Now, to get the locus, we will replace h by x and k by y. We get,

\[{{y}^{2}}-by-2ax+6ab=0\]

So, the locus of the midpoint of chord passing through \[\left( 3b,b \right)\] is \[{{y}^{2}}-by-2ax+6ab=0\].

Note: In these types of questions, we can directly write the equation of chord with respect to mid-point say \[\left( {{x}_{1}},{{y}_{1}} \right)\] which is \[\left( y-{{y}_{1}} \right)=\dfrac{2a}{{{y}_{1}}}\left( x-{{x}_{1}} \right)\] and put \[\left( h,k \right)\] in place of \[\left( {{x}_{1}},{{y}_{1}} \right)\] and point through which chord in passing [here (3b,b)] in place of \[\left( x,y \right)\] to get locus of midpoint of chord.

Last updated date: 22nd Sep 2023

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