# Two tangents to a parabola intercept on a fixed tangent segment whose product is a constant. Prove that the locus of their point of intersection is a straight line.

Last updated date: 21st Mar 2023

•

Total views: 307.5k

•

Views today: 3.86k

Answer

Verified

307.5k+ views

Hint: The general equation of tangent at \[\left( a{{t}^{2}},2at \right)\]is given by \[ty=x+a{{t}^{2}}\], where \[t\] is a parameter .

The point of intersection of tangents at \[A\left( at_{1}^{2},2a{{t}_{1}} \right)\]and \[B\left( at_{2}^{2},2a{{t}_{2}} \right)\] is given by \[\left( a{{t}_{1}}{{t}_{2}},a\left( {{t}_{1}}+{{t}_{2}} \right) \right)\], where \[{{t}_{1}}\] and \[{{t}_{2}}\] are parameters.

We will consider the equation of the parabola to be \[{{y}^{2}}=4ax\].

We will consider two points \[A\left( at_{1}^{2},2a{{t}_{1}} \right)\] and \[B\left( at_{2}^{2},2a{{t}_{2}} \right)\] on the parabola , where \[{{t}_{1}}\] and \[{{t}_{2}}\] are parameters.

Now , we will find the equation of tangents at these points.

Now, we know the general equation of tangent at \[\left( a{{t}^{2}},2at \right)\] is given by \[ty=x+a{{t}^{2}}\], where \[t\] is a parameter .

So , the equation of tangent at \[A\left( at_{1}^{2},2a{{t}_{1}} \right)\] is given by substituting \[{{t}_{1}}\] in place of \[t\] in the general equation of tangent .

On substituting \[{{t}_{1}}\] in place of \[t\] in the general equation of tangent , we get

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

And equation of tangent at \[B\left( at_{2}^{2},2a{{t}_{2}} \right)\] is given as

\[{{t}_{2}}y=x+at_{2}^{2}....\left( ii \right)\]

Now, we need to find the locus of the point of intersection of \[\left( i \right)\]and \[\left( ii \right)\].

Let the point of intersection be \[M\left( h,k \right)\].

Now, from equation\[\left( i \right)\], we have

\[y{{t}_{1}}=x+at_{1}^{2}\]

\[\Rightarrow x={{t}_{1}}\left( y-a{{t}_{1}} \right).....\left( iii \right)\]

We will substitute the value of \[x\] from equation \[(iii)\] in equation \[\left( ii \right)\].

On substituting value of \[x\] from equation\[(iii)\] in equation \[\left( ii \right)\], we get,

\[y{{t}_{2}}={{t}_{1}}y-at_{1}^{2}+at_{2}^{2}\]

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

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

Substituting \[y=a\left( {{t}_{1}}+{{t}_{2}} \right)\]in \[\left( iii \right)\], we get

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

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

So, the point of intersection of tangents \[\left( i \right)\] and \[\left( ii \right)\] is \[\left( a{{t}_{1}}{{t}_{2}},a\left( {{t}_{1}}+{{t}_{2}} \right) \right)\].

Comparing it with\[M\left( h,k \right)\], we get

\[h=a{{t}_{1}}{{t}_{2}}\]

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

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

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

Now, we will consider the point of contact of the fixed tangent be \[C\left( a{{t}^{2}},2at \right)\].

So , the point of intersection of tangent at \[A\] and tangent at \[C\] is \[D\left( at{{t}_{1}},a\left( t+{{t}_{1}} \right) \right)\].

And the point of intersection of tangent at \[B\] and tangent at \[C\] is \[E\left( at{{t}_{2}},a\left( t+{{t}_{2}} \right) \right)\].

Now, in the question it is given that the product of intercept on the fixed tangent is constant.

So, \[CD\times CE=C\]

\[\Rightarrow \sqrt{{{\left( a{{t}^{2}}-at{{t}_{1}} \right)}^{2}}+{{\left( 2at-a{{t}_{1}}-at \right)}^{2}}}\times \sqrt{{{\left( a{{t}^{2}}-at{{t}_{2}} \right)}^{2}}+{{\left( 2at-a{{t}_{2}}-at \right)}^{2}}}=C\]

\[\Rightarrow \sqrt{{{a}^{2}}{{t}^{2}}{{\left( t-{{t}_{1}} \right)}^{2}}+{{a}^{2}}{{\left( t-{{t}_{1}} \right)}^{2}}}\times \sqrt{{{a}^{2}}{{t}^{2}}{{\left( t-{{t}_{2}} \right)}^{2}}+{{a}^{2}}{{\left( t-{{t}_{2}} \right)}^{2}}}=C\]

\[\Rightarrow a\left( t-{{t}_{1}} \right)\sqrt{{{t}^{2}}+1}\times a\left( t-{{t}_{2}} \right)\sqrt{{{t}^{2}}+1}=C\]

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

\[\Rightarrow \left( t-{{t}_{1}} \right)\left( t-{{t}_{2}} \right)={{C}_{1}}\] (say)

\[\Rightarrow {{t}^{2}}-\left( {{t}_{1}}+{{t}_{2}} \right)+{{t}_{1}}{{t}_{2}}={{C}_{1}}....\left( vi \right)\]

Substituting \[\left( iv \right)\]and \[\left( v \right)\]in \[\left( vi \right)\], we get

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

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

Now, to find the locus of \[M\left( h,k \right)\], we will substitute \[(x,y)\] in place of \[(h,k)\] in equation\[(vii)\].

So , the locus of \[M\left( h,k \right)\] is given as

\[x-y=a\left( {{C}_{1}}-{{t}^{2}} \right)\] which represents a straight line.

Note: : While simplifying the equations , please make sure that sign mistakes do not occur. These mistakes are very common and can cause confusions while solving. Ultimately the answer becomes wrong. So, sign conventions should be carefully taken .

The point of intersection of tangents at \[A\left( at_{1}^{2},2a{{t}_{1}} \right)\]and \[B\left( at_{2}^{2},2a{{t}_{2}} \right)\] is given by \[\left( a{{t}_{1}}{{t}_{2}},a\left( {{t}_{1}}+{{t}_{2}} \right) \right)\], where \[{{t}_{1}}\] and \[{{t}_{2}}\] are parameters.

We will consider the equation of the parabola to be \[{{y}^{2}}=4ax\].

We will consider two points \[A\left( at_{1}^{2},2a{{t}_{1}} \right)\] and \[B\left( at_{2}^{2},2a{{t}_{2}} \right)\] on the parabola , where \[{{t}_{1}}\] and \[{{t}_{2}}\] are parameters.

Now , we will find the equation of tangents at these points.

Now, we know the general equation of tangent at \[\left( a{{t}^{2}},2at \right)\] is given by \[ty=x+a{{t}^{2}}\], where \[t\] is a parameter .

So , the equation of tangent at \[A\left( at_{1}^{2},2a{{t}_{1}} \right)\] is given by substituting \[{{t}_{1}}\] in place of \[t\] in the general equation of tangent .

On substituting \[{{t}_{1}}\] in place of \[t\] in the general equation of tangent , we get

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

And equation of tangent at \[B\left( at_{2}^{2},2a{{t}_{2}} \right)\] is given as

\[{{t}_{2}}y=x+at_{2}^{2}....\left( ii \right)\]

Now, we need to find the locus of the point of intersection of \[\left( i \right)\]and \[\left( ii \right)\].

Let the point of intersection be \[M\left( h,k \right)\].

Now, from equation\[\left( i \right)\], we have

\[y{{t}_{1}}=x+at_{1}^{2}\]

\[\Rightarrow x={{t}_{1}}\left( y-a{{t}_{1}} \right).....\left( iii \right)\]

We will substitute the value of \[x\] from equation \[(iii)\] in equation \[\left( ii \right)\].

On substituting value of \[x\] from equation\[(iii)\] in equation \[\left( ii \right)\], we get,

\[y{{t}_{2}}={{t}_{1}}y-at_{1}^{2}+at_{2}^{2}\]

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

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

Substituting \[y=a\left( {{t}_{1}}+{{t}_{2}} \right)\]in \[\left( iii \right)\], we get

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

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

So, the point of intersection of tangents \[\left( i \right)\] and \[\left( ii \right)\] is \[\left( a{{t}_{1}}{{t}_{2}},a\left( {{t}_{1}}+{{t}_{2}} \right) \right)\].

Comparing it with\[M\left( h,k \right)\], we get

\[h=a{{t}_{1}}{{t}_{2}}\]

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

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

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

Now, we will consider the point of contact of the fixed tangent be \[C\left( a{{t}^{2}},2at \right)\].

So , the point of intersection of tangent at \[A\] and tangent at \[C\] is \[D\left( at{{t}_{1}},a\left( t+{{t}_{1}} \right) \right)\].

And the point of intersection of tangent at \[B\] and tangent at \[C\] is \[E\left( at{{t}_{2}},a\left( t+{{t}_{2}} \right) \right)\].

Now, in the question it is given that the product of intercept on the fixed tangent is constant.

So, \[CD\times CE=C\]

\[\Rightarrow \sqrt{{{\left( a{{t}^{2}}-at{{t}_{1}} \right)}^{2}}+{{\left( 2at-a{{t}_{1}}-at \right)}^{2}}}\times \sqrt{{{\left( a{{t}^{2}}-at{{t}_{2}} \right)}^{2}}+{{\left( 2at-a{{t}_{2}}-at \right)}^{2}}}=C\]

\[\Rightarrow \sqrt{{{a}^{2}}{{t}^{2}}{{\left( t-{{t}_{1}} \right)}^{2}}+{{a}^{2}}{{\left( t-{{t}_{1}} \right)}^{2}}}\times \sqrt{{{a}^{2}}{{t}^{2}}{{\left( t-{{t}_{2}} \right)}^{2}}+{{a}^{2}}{{\left( t-{{t}_{2}} \right)}^{2}}}=C\]

\[\Rightarrow a\left( t-{{t}_{1}} \right)\sqrt{{{t}^{2}}+1}\times a\left( t-{{t}_{2}} \right)\sqrt{{{t}^{2}}+1}=C\]

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

\[\Rightarrow \left( t-{{t}_{1}} \right)\left( t-{{t}_{2}} \right)={{C}_{1}}\] (say)

\[\Rightarrow {{t}^{2}}-\left( {{t}_{1}}+{{t}_{2}} \right)+{{t}_{1}}{{t}_{2}}={{C}_{1}}....\left( vi \right)\]

Substituting \[\left( iv \right)\]and \[\left( v \right)\]in \[\left( vi \right)\], we get

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

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

Now, to find the locus of \[M\left( h,k \right)\], we will substitute \[(x,y)\] in place of \[(h,k)\] in equation\[(vii)\].

So , the locus of \[M\left( h,k \right)\] is given as

\[x-y=a\left( {{C}_{1}}-{{t}^{2}} \right)\] which represents a straight line.

Note: : While simplifying the equations , please make sure that sign mistakes do not occur. These mistakes are very common and can cause confusions while solving. Ultimately the answer becomes wrong. So, sign conventions should be carefully taken .

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

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

Tropic of Cancer passes through how many states? Name them.

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

Name the Largest and the Smallest Cell in the Human Body ?