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Hint: Write the equation of tangent, get points \[Q\] and \[P\] and then use \[x=\dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n}\] and \[y=\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n}\].

We are given a point \[P\] on the parabola \[{{y}^{2}}-2y-4x+5=0\] such that tangent on the parabola at \[P\] intersect the directrix at \[Q\].

We have to find the locus of \[R\] which divides \[QP\] externally in the ratio \[\dfrac{1}{2}:1\].

First of all, we will convert the parabola into its standard form

We take, \[{{y}^{2}}-2y-4x+5=0\]

Or, \[{{y}^{2}}-2y=4x-5\]

Adding \[1\] on both sides,

We get, \[{{y}^{2}}-2y+1=4x-5+1\]

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

Therefore, we get \[{{\left( y-1 \right)}^{2}}=4x-4\]

Or, \[{{\left( y-1 \right)}^{2}}=4\left( x-1 \right)\]

Let \[y-1=Y\] and \[x-1=X\]

So, we get parabola \[{{Y}^{2}}=4X\] which is the standard form of a parabola.

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

By comparing parabola \[{{Y}^{2}}=4X\] with general parabola \[{{y}^{2}}=4ax\],

We get \[4a=4\]

Therefore, we get \[a=1\]

Now, we know that any general point on the parabola \[{{y}^{2}}=4ax\] is \[\left( a{{t}^{2}},2at \right)\]

Since, \[a=1\]

We get, \[P\left( X,Y \right)=P\left( {{t}^{2}},2t \right)\]

We know that tangent at point \[\left( a{{t}^{2}},2at \right)\] is given by \[ty=x+a{{t}^{2}}\]

Since \[a=1\], we get equation of tangent at \[P\left( {{t}^{2}},2t \right)\] as \[tY=X+{{t}^{2}}....\left( i \right)\]

We know that equation of directrix is

\[X=-a\]

Therefore, we get \[X=-1\]

Or, \[X+1=0\]

Since we know that \[Q\] is a point of intersection of tangents and directrix is \[X=-1\].

Therefore, we will put \[X=-1\] in the equation \[\left( i \right)\] to find the \[Y\] coordinate.

So, \[tY=X+{{t}^{2}}\]

\[\Rightarrow tY=-1+{{t}^{2}}\]

\[Y=\dfrac{{{t}^{2}}-1}{t}\]

Therefore, we get a point \[Q\left( X,Y \right)=\left( -1,\dfrac{{{t}^{2}}-1}{t} \right)\]

Let the point \[R\] be \[\left( h,k \right)\] which divides \[QP\] externally in the ratio \[\dfrac{1}{2}:1\]

We know that if \[R\left( x,y \right)\] divides \[Q\left( {{x}_{1}},{{y}_{1}} \right)\] and \[P\left( {{x}_{2}},{{y}_{2}} \right)\]in the ratio \[m:n\]

Then, \[x=\dfrac{m\left( {{x}_{2}} \right)-n\left( {{x}_{1}} \right)}{\left( m-n \right)}\]

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

Here, we have \[R\left( x,y \right)=\left( h,k \right)\]

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

\[P\left( {{x}_{2}},{{y}_{2}} \right)=\left( {{t}^{2}},2t \right)\]

\[m=\dfrac{1}{2},n=1\]

So, we get \[h=\dfrac{\dfrac{1}{2}\left( {{t}^{2}} \right)-1\left( -1 \right)}{\dfrac{1}{2}-1}\]

\[h=\dfrac{\dfrac{{{t}^{2}}}{2}+1}{\dfrac{-1}{2}}\]

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

Or \[{{t}^{2}}=-2-h....\left( ii \right)\]

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

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

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

\[k=\dfrac{-2}{t}\]

By squaring both the sides,

We get, \[{{k}^{2}}=\dfrac{4}{{{t}^{2}}}\]

Now, by putting the value of \[{{t}^{2}}\] from equation \[\left( ii \right)\]

We get \[{{k}^{2}}=\dfrac{4}{\left( -2-h \right)}\]

By cross multiplying, we get

\[\Rightarrow -{{k}^{2}}\left( h+2 \right)=4\]

To get the locus, we will replace \[h\]by \[X\] and \[k\]by \[Y\].

We get, \[-{{Y}^{2}}\left( X+2 \right)=4\]

As we had assumed that \[X=x-1\] and \[Y=y-1\]

We get, \[-{{\left( y-1 \right)}^{2}}\left( x-1+2 \right)=4\]

\[-{{\left( y-1 \right)}^{2}}\left( x+1 \right)=4\]

Or, \[{{\left( y-1 \right)}^{2}}\left( x+1 \right)+4=0\]

Note: Always convert the given parabola into standard parabola \[{{y}^{2}}=4ax\] and then use the general equation of tangents, normals etc. Also, students often forget to convert \[X\] to \[x\] and \[Y\] to \[y\] and get wrong answers. So this step must be kept in mind.

We are given a point \[P\] on the parabola \[{{y}^{2}}-2y-4x+5=0\] such that tangent on the parabola at \[P\] intersect the directrix at \[Q\].

We have to find the locus of \[R\] which divides \[QP\] externally in the ratio \[\dfrac{1}{2}:1\].

First of all, we will convert the parabola into its standard form

We take, \[{{y}^{2}}-2y-4x+5=0\]

Or, \[{{y}^{2}}-2y=4x-5\]

Adding \[1\] on both sides,

We get, \[{{y}^{2}}-2y+1=4x-5+1\]

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

Therefore, we get \[{{\left( y-1 \right)}^{2}}=4x-4\]

Or, \[{{\left( y-1 \right)}^{2}}=4\left( x-1 \right)\]

Let \[y-1=Y\] and \[x-1=X\]

So, we get parabola \[{{Y}^{2}}=4X\] which is the standard form of a parabola.

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

By comparing parabola \[{{Y}^{2}}=4X\] with general parabola \[{{y}^{2}}=4ax\],

We get \[4a=4\]

Therefore, we get \[a=1\]

Now, we know that any general point on the parabola \[{{y}^{2}}=4ax\] is \[\left( a{{t}^{2}},2at \right)\]

Since, \[a=1\]

We get, \[P\left( X,Y \right)=P\left( {{t}^{2}},2t \right)\]

We know that tangent at point \[\left( a{{t}^{2}},2at \right)\] is given by \[ty=x+a{{t}^{2}}\]

Since \[a=1\], we get equation of tangent at \[P\left( {{t}^{2}},2t \right)\] as \[tY=X+{{t}^{2}}....\left( i \right)\]

We know that equation of directrix is

\[X=-a\]

Therefore, we get \[X=-1\]

Or, \[X+1=0\]

Since we know that \[Q\] is a point of intersection of tangents and directrix is \[X=-1\].

Therefore, we will put \[X=-1\] in the equation \[\left( i \right)\] to find the \[Y\] coordinate.

So, \[tY=X+{{t}^{2}}\]

\[\Rightarrow tY=-1+{{t}^{2}}\]

\[Y=\dfrac{{{t}^{2}}-1}{t}\]

Therefore, we get a point \[Q\left( X,Y \right)=\left( -1,\dfrac{{{t}^{2}}-1}{t} \right)\]

Let the point \[R\] be \[\left( h,k \right)\] which divides \[QP\] externally in the ratio \[\dfrac{1}{2}:1\]

We know that if \[R\left( x,y \right)\] divides \[Q\left( {{x}_{1}},{{y}_{1}} \right)\] and \[P\left( {{x}_{2}},{{y}_{2}} \right)\]in the ratio \[m:n\]

Then, \[x=\dfrac{m\left( {{x}_{2}} \right)-n\left( {{x}_{1}} \right)}{\left( m-n \right)}\]

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

Here, we have \[R\left( x,y \right)=\left( h,k \right)\]

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

\[P\left( {{x}_{2}},{{y}_{2}} \right)=\left( {{t}^{2}},2t \right)\]

\[m=\dfrac{1}{2},n=1\]

So, we get \[h=\dfrac{\dfrac{1}{2}\left( {{t}^{2}} \right)-1\left( -1 \right)}{\dfrac{1}{2}-1}\]

\[h=\dfrac{\dfrac{{{t}^{2}}}{2}+1}{\dfrac{-1}{2}}\]

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

Or \[{{t}^{2}}=-2-h....\left( ii \right)\]

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

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

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

\[k=\dfrac{-2}{t}\]

By squaring both the sides,

We get, \[{{k}^{2}}=\dfrac{4}{{{t}^{2}}}\]

Now, by putting the value of \[{{t}^{2}}\] from equation \[\left( ii \right)\]

We get \[{{k}^{2}}=\dfrac{4}{\left( -2-h \right)}\]

By cross multiplying, we get

\[\Rightarrow -{{k}^{2}}\left( h+2 \right)=4\]

To get the locus, we will replace \[h\]by \[X\] and \[k\]by \[Y\].

We get, \[-{{Y}^{2}}\left( X+2 \right)=4\]

As we had assumed that \[X=x-1\] and \[Y=y-1\]

We get, \[-{{\left( y-1 \right)}^{2}}\left( x-1+2 \right)=4\]

\[-{{\left( y-1 \right)}^{2}}\left( x+1 \right)=4\]

Or, \[{{\left( y-1 \right)}^{2}}\left( x+1 \right)+4=0\]

Note: Always convert the given parabola into standard parabola \[{{y}^{2}}=4ax\] and then use the general equation of tangents, normals etc. Also, students often forget to convert \[X\] to \[x\] and \[Y\] to \[y\] and get wrong answers. So this step must be kept in mind.

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