# The normal at any point \[P\] meets the axis in \[G\] and the tangent at the vertex in \[{{G}^{'}}\]. If \[A\] be the vertex and the rectangle \[AGQ{{G}^{'}}\] be completed, prove that the equation of locus of \[Q\] is

\[{{x}^{3}}=2a{{x}^{2}}+a{{y}^{2}}\]

Answer

Verified

383.1k+ views

Hint:Here we use the property of a rectangle that the mid-points of the diagonals of a rectangle are the same .

Let’s consider the equation of the parabola to be \[{{y}^{2}}=4ax\]. So, its vertex is \[A(0,0)\].

We know , any point on the parabola , given by the equation \[{{y}^{2}}=4ax\], can be written as \[P\left( a{{t}^{2}},2at \right)\].

Now , we will find the equation of normal to the parabola at \[P\].

We know , the equation of the normal to the parabola in parametric form is given as

\[y=-tx+2at+a{{t}^{3}}\]

So, the equation of normal at \[P\] can be written as

\[y=-tx+2at+a{{t}^{3}}....\left( i \right)\]

Now , we know , the axis of the parabola is \[y=0\].

We need to find the point of intersection of the normal to the parabola and the axis of the parabola .

To find the point of intersection of normal and the axis , we will substitute \[y=0\] in equation \[\left( i \right)\].

On substituting \[y=0\] in equation \[\left( i \right)\] , we get

\[0=-tx+2at+a{{t}^{3}}\]

\[\Rightarrow x=2a+a{{t}^{2}}\]

So , the coordinates of \[G\] are \[\left( 2a+a{{t}^{2}},0 \right)\].

Now , we know , tangent to the parabola at vertex is given by the equation \[x=0\].

So , the point of intersection of normal and tangent at vertex can be found by substituting \[x=0\]in \[\left( i \right)\].

On substituting \[x=0\]in \[\left( i \right)\], we get

\[y=2at+a{{t}^{3}}\]

So, \[{{G}^{'}}=\left( 0,2at+a{{t}^{3}} \right)\]

Now , since we need to find the locus of \[Q\], let \[Q=\left( h,k \right)\].

Now , it is given that \[AGQ{{G}^{'}}\]is a rectangle.

We know the diagonals of a rectangle bisect each other.

So , the midpoint of the diagonals \[AQ\] and \[G{{G}^{'}}\] is the same.

Now, we know that the coordinates of the midpoint of the line joining two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is given as: \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]

So, midpoint of \[AQ=\left( \dfrac{0+h}{2},\dfrac{0+k}{2} \right)\]

\[=\left( \dfrac{h}{2},\dfrac{k}{2} \right)....\left( ii \right)\]

Midpoint of \[G{{G}^{'}}=\left( \dfrac{2a+a{{t}^{2}}+0}{2},\dfrac{0+2at+a{{t}^{3}}}{2} \right)\]

\[=\left( \dfrac{2a+a{{t}^{2}}}{2},\dfrac{2at+a{{t}^{3}}}{2} \right)....\left( iii \right)\]

Comparing \[\left( ii \right)\] with \[\left( iii \right)\],

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

\[\dfrac{k}{2}=\dfrac{2at+a{{t}^{3}}}{2}\Rightarrow k=2at+a{{t}^{3}}....\left( v \right)\]

On dividing \[\left( v \right)\]by \[\left( iv \right)\], we get

\[\dfrac{k}{h}=\dfrac{2at+a{{t}^{3}}}{2a+a{{t}^{2}}}\]

\[\Rightarrow \dfrac{k}{h}=t\]

Now, let’s substitute \[t=\dfrac{k}{h}\]in \[\left( iv \right)\]. We get ,

\[h=2a+a{{\left( \dfrac{k}{h} \right)}^{2}}\]

\[\Rightarrow h=\dfrac{2a{{h}^{2}}+a{{k}^{2}}}{{{h}^{2}}}\]

Or, \[{{h}^{3}}=2a{{h}^{2}}+a{{k}^{2}}........\]equation\[(vi)\]

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

So, the locus of \[Q\left( h,k \right)\] is \[{{x}^{3}}=2a{{x}^{2}}+a{{y}^{2}}\]

Note: Vertex of \[{{y}^{2}}=4ax\]is \[\left( 0,0 \right)\].

Tangent at vertex is \[x=0\]

Equation of the axis is \[y=0\].

Students generally get confused between the equation of tangent at vertex and the equation of axis.

Let’s consider the equation of the parabola to be \[{{y}^{2}}=4ax\]. So, its vertex is \[A(0,0)\].

We know , any point on the parabola , given by the equation \[{{y}^{2}}=4ax\], can be written as \[P\left( a{{t}^{2}},2at \right)\].

Now , we will find the equation of normal to the parabola at \[P\].

We know , the equation of the normal to the parabola in parametric form is given as

\[y=-tx+2at+a{{t}^{3}}\]

So, the equation of normal at \[P\] can be written as

\[y=-tx+2at+a{{t}^{3}}....\left( i \right)\]

Now , we know , the axis of the parabola is \[y=0\].

We need to find the point of intersection of the normal to the parabola and the axis of the parabola .

To find the point of intersection of normal and the axis , we will substitute \[y=0\] in equation \[\left( i \right)\].

On substituting \[y=0\] in equation \[\left( i \right)\] , we get

\[0=-tx+2at+a{{t}^{3}}\]

\[\Rightarrow x=2a+a{{t}^{2}}\]

So , the coordinates of \[G\] are \[\left( 2a+a{{t}^{2}},0 \right)\].

Now , we know , tangent to the parabola at vertex is given by the equation \[x=0\].

So , the point of intersection of normal and tangent at vertex can be found by substituting \[x=0\]in \[\left( i \right)\].

On substituting \[x=0\]in \[\left( i \right)\], we get

\[y=2at+a{{t}^{3}}\]

So, \[{{G}^{'}}=\left( 0,2at+a{{t}^{3}} \right)\]

Now , since we need to find the locus of \[Q\], let \[Q=\left( h,k \right)\].

Now , it is given that \[AGQ{{G}^{'}}\]is a rectangle.

We know the diagonals of a rectangle bisect each other.

So , the midpoint of the diagonals \[AQ\] and \[G{{G}^{'}}\] is the same.

Now, we know that the coordinates of the midpoint of the line joining two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is given as: \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]

So, midpoint of \[AQ=\left( \dfrac{0+h}{2},\dfrac{0+k}{2} \right)\]

\[=\left( \dfrac{h}{2},\dfrac{k}{2} \right)....\left( ii \right)\]

Midpoint of \[G{{G}^{'}}=\left( \dfrac{2a+a{{t}^{2}}+0}{2},\dfrac{0+2at+a{{t}^{3}}}{2} \right)\]

\[=\left( \dfrac{2a+a{{t}^{2}}}{2},\dfrac{2at+a{{t}^{3}}}{2} \right)....\left( iii \right)\]

Comparing \[\left( ii \right)\] with \[\left( iii \right)\],

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

\[\dfrac{k}{2}=\dfrac{2at+a{{t}^{3}}}{2}\Rightarrow k=2at+a{{t}^{3}}....\left( v \right)\]

On dividing \[\left( v \right)\]by \[\left( iv \right)\], we get

\[\dfrac{k}{h}=\dfrac{2at+a{{t}^{3}}}{2a+a{{t}^{2}}}\]

\[\Rightarrow \dfrac{k}{h}=t\]

Now, let’s substitute \[t=\dfrac{k}{h}\]in \[\left( iv \right)\]. We get ,

\[h=2a+a{{\left( \dfrac{k}{h} \right)}^{2}}\]

\[\Rightarrow h=\dfrac{2a{{h}^{2}}+a{{k}^{2}}}{{{h}^{2}}}\]

Or, \[{{h}^{3}}=2a{{h}^{2}}+a{{k}^{2}}........\]equation\[(vi)\]

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

So, the locus of \[Q\left( h,k \right)\] is \[{{x}^{3}}=2a{{x}^{2}}+a{{y}^{2}}\]

Note: Vertex of \[{{y}^{2}}=4ax\]is \[\left( 0,0 \right)\].

Tangent at vertex is \[x=0\]

Equation of the axis is \[y=0\].

Students generally get confused between the equation of tangent at vertex and the equation of axis.

Recently Updated Pages

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

What is 1 divided by 0 class 8 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

How many crores make 10 million class 7 maths CBSE

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

State the laws of reflection of light

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE