Prove that the locus of the point of intersection of two tangents which intercept a given distance \[4c\] on the tangent at the vertex is an equal parabola.
VerifiedHint: Equation of tangent at vertex \[{{y}^{2}}=4ax\] is given as \[x=0\].
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.
Complete step-by-step answer:
We will consider the equation of parabola to be \[{{y}^{2}}=4ax\].
We know, the equation of tangent at vertex is given as \[x=0......\]equation\[(i)\].
Now, we will consider two points on the parabola , given by \[A\left( at_{1}^{2},2a{{t}_{1}} \right)\] and \[B\left( at_{2}^{2},2a{{t}_{2}} \right)\], 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
\[{{t}_{1}}y=x+at_{1}^{2}.....\left( ii \right)\]
Similarly, the equation of tangent at \[B\left( at_{2}^{2},2a{{t}_{2}} \right)\] is given as
\[{{t}_{2}}y=x+at_{2}^{2}.....\left( iii \right)\]
Now , to find the intercept of \[\left( ii \right)\]on \[\left( i \right)\], we will substitute \[x=0\] in equation\[\left( ii \right)\].
On substituting\[x=0\] in equation\[\left( ii \right)\], we get
\[{{t}_{1}}y=at_{1}^{2}\]
Or, \[y=a{{t}_{1}}\]
So, the point of intersection is \[\left( 0,a{{t}_{1}} \right)\] and length of intercept is \[a{{t}_{1}}\].
Now, to find the intercept of \[\left( iii \right)\]on \[\left( i \right)\], we will substitute \[x=0\] in equation\[\left( iii \right)\].
On substituting\[x=0\] in equation\[\left( iii \right)\], we get
\[{{t}_{2}}y=at_{2}^{2}\]
\[\Rightarrow y=a{{t}_{2}}\]
So , the point of intersection is \[\left( 0,a{{t}_{2}} \right)\] and length of intercept is \[a{{t}_{2}}\].
Now, in the question it is given that the tangents \[\left( ii \right)\]and \[\left( iii \right)\] intercept a distance \[4c\] on the tangent at vertex.
So, \[a({{t}_{1}}-{{t}_{2}})=4c......(iv)\]
Now, we need to find the locus of points of intersection of tangents \[\left( ii \right)\] and \[\left( iii \right)\].
Let this point be \[M\left( h,k \right)\].
Now, from equation\[\left( ii \right)\], we have
\[y{{t}_{1}}=x+at_{1}^{2}\]
\[\Rightarrow x={{t}_{1}}\left( y-a{{t}_{1}} \right).....\left( v \right)\]
We will substitute the value of \[x\] from equation \[(v)\] in equation \[\left( iii \right)\].
On substituting value of \[x\] from equation\[(v)\] in equation \[\left( iii \right)\], 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 \[\left( v \right)\], 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 \[\left( ii \right)\]and \[\left( iii \right)\]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
\[h=a{{t}_{1}}{{t}_{2}}\]
\[\Rightarrow \dfrac{h}{a}={{t}_{1}}{{t}_{2}}.....\left( vi \right)\]
And \[k=a\left( {{t}_{1}}+{{t}_{2}} \right)\]
\[\Rightarrow \dfrac{k}{a}={{t}_{1}}+{{t}_{2}}.....\left( vii \right)\]
Now, from \[\left( vii \right)\], we have
\[k=a\left( {{t}_{1}}+{{t}_{2}} \right)\]
\[\Rightarrow {{k}^{2}}={{a}^{2}}{{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}\]
\[\Rightarrow {{k}^{2}}={{a}^{2}}\left( {{\left( {{t}_{1}}-{{t}_{2}} \right)}^{2}}+4{{t}_{1}}{{t}_{2}} \right)\]
Now, from \[\left( iv \right)\], we have
\[\left( {{t}_{1}}-{{t}_{2}} \right)=\dfrac{4c}{a}\]
So, \[{{k}^{2}}={{a}^{2}}\left( {{\left( \dfrac{4c}{a} \right)}^{2}}+4\dfrac{h}{a} \right)\]
\[\Rightarrow {{k}^{2}}={{a}^{2}}\left( \dfrac{16{{c}^{2}}}{{{a}^{2}}}+\dfrac{4h}{a} \right)\]
\[\Rightarrow {{k}^{2}}=16{{c}^{2}}+4ah.........(viii)\]
Now, to find the locus of \[M\left( h,k \right)\], we will substitute \[(x,y)\] in place of \[(h,k)\] in equation \[(viii)\].
So , the locus of \[M\left( h,k \right)\] is given as
\[{{y}^{2}}=16{{c}^{2}}+4ax\]
\[\Rightarrow {{y}^{2}}=4a\left( x+\dfrac{4{{c}^{2}}}{a} \right)\] which represents a parabola.
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.
