Question

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 1

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.