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

Last updated date: 27th Mar 2023

•

Total views: 307.2k

•

Views today: 3.84k

Answer

Verified

307.2k+ views

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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE