# Prove that the locus of the middle point of a portion of a normal intersected between the curve and the axis is a parabola whose vertex is the focus and whose latus rectum is one quarter of the original parabola.

Answer

Verified

363.9k+ views

Hint: The equation of normal in parametric form is given as \[y=-tx+2at+a{{t}^{3}}\], where the normal is drawn at a point \[P\] with parameter \[t\].

First of all , let’s take the equation of the parabola to be \[{{y}^{2}}=4ax\].

Coordinates of any point on the parabola in parametric form is given as \[P\left( a{{t}^{2}},2at \right)\].

We know, equation of normal in parametric form , where the normal is drawn at a point \[P\] with parameter \[t\], is given as

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

Also, the equation of axis of the parabola is

\[y=0.....\left( ii \right)\]

To find the points of intersection of \[\left( i \right)\]and\[\left( ii \right)\], we substitute \[y=0\] in \[\left( i \right)\].

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

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

So , the point of intersection of the axis and the normal is \[\left( 2a+a{{t}^{2}},0 \right)\].

Now , let the midpoint of intercepted portion be \[\left( h,k \right).....\left( iii \right)\]

But we also know that the extremities of the intercepted portion are \[\left( a{{t}^{2}},2at \right)\] and \[\left( 2a+a{{t}^{2}},0 \right)\].

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, the coordinates of the midpoint are

\[\left( \dfrac{a{{t}^{2}}+2a+a{{t}^{2}}}{2},\dfrac{2at+0}{2} \right)\]

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

So, from \[\left( iii \right)\]and \[\left( iv \right)\], we can say

\[h=a{{t}^{2}}+a....\left( v \right)\], \[k=at\]

Now, \[k=at\]\[\Rightarrow t=\dfrac{k}{a}...\left( vi \right)\]

Substituting \[t=\dfrac{k}{a}\]in \[\left( v \right)\], we get

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

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

\[\Rightarrow {{k}^{2}}=ah-{{a}^{2}}........\] equation\[(vii)\]

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

Or, \[{{y}^{2}}=a\left( x-a \right)....\left( viii \right)\] is the equation of the locus.

Now , we know the length of the latus rectum of \[{{y}^{2}}=4ax\] is \[4a\].

Now , the length of latus rectum of parabola given by equation \[\left( vii \right)\]is \[4\times \dfrac{a}{4}=a\]

Also, the vertex of \[\left( viii \right)\] is \[\left( a,0 \right)\] which is the focus of \[{{y}^{2}}=4ax\]

Hence , the length of the latus rectum of the locus is one quarter of the original parabola and the vertex of the locus is the focus of the original parabola .

Note: Length of latus rectum of parabola \[{{y}^{2}}=4ax\] is equal to \[4a\] and not \[a\].

Focus of parabola \[{{y}^{2}}=4ax\] is \[\left( a,0 \right)\] and not \[(4a,0)\].

Students generally get confused and make mistakes which results in wrong answers. So , such mistakes should be avoided .

First of all , let’s take the equation of the parabola to be \[{{y}^{2}}=4ax\].

Coordinates of any point on the parabola in parametric form is given as \[P\left( a{{t}^{2}},2at \right)\].

We know, equation of normal in parametric form , where the normal is drawn at a point \[P\] with parameter \[t\], is given as

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

Also, the equation of axis of the parabola is

\[y=0.....\left( ii \right)\]

To find the points of intersection of \[\left( i \right)\]and\[\left( ii \right)\], we substitute \[y=0\] in \[\left( i \right)\].

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

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

So , the point of intersection of the axis and the normal is \[\left( 2a+a{{t}^{2}},0 \right)\].

Now , let the midpoint of intercepted portion be \[\left( h,k \right).....\left( iii \right)\]

But we also know that the extremities of the intercepted portion are \[\left( a{{t}^{2}},2at \right)\] and \[\left( 2a+a{{t}^{2}},0 \right)\].

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, the coordinates of the midpoint are

\[\left( \dfrac{a{{t}^{2}}+2a+a{{t}^{2}}}{2},\dfrac{2at+0}{2} \right)\]

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

So, from \[\left( iii \right)\]and \[\left( iv \right)\], we can say

\[h=a{{t}^{2}}+a....\left( v \right)\], \[k=at\]

Now, \[k=at\]\[\Rightarrow t=\dfrac{k}{a}...\left( vi \right)\]

Substituting \[t=\dfrac{k}{a}\]in \[\left( v \right)\], we get

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

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

\[\Rightarrow {{k}^{2}}=ah-{{a}^{2}}........\] equation\[(vii)\]

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

Or, \[{{y}^{2}}=a\left( x-a \right)....\left( viii \right)\] is the equation of the locus.

Now , we know the length of the latus rectum of \[{{y}^{2}}=4ax\] is \[4a\].

Now , the length of latus rectum of parabola given by equation \[\left( vii \right)\]is \[4\times \dfrac{a}{4}=a\]

Also, the vertex of \[\left( viii \right)\] is \[\left( a,0 \right)\] which is the focus of \[{{y}^{2}}=4ax\]

Hence , the length of the latus rectum of the locus is one quarter of the original parabola and the vertex of the locus is the focus of the original parabola .

Note: Length of latus rectum of parabola \[{{y}^{2}}=4ax\] is equal to \[4a\] and not \[a\].

Focus of parabola \[{{y}^{2}}=4ax\] is \[\left( a,0 \right)\] and not \[(4a,0)\].

Students generally get confused and make mistakes which results in wrong answers. So , such mistakes should be avoided .

Last updated date: 27th Sep 2023

•

Total views: 363.9k

•

Views today: 5.63k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

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

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

How many millions make a billion class 6 maths CBSE

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

Number of Prime between 1 to 100 is class 6 maths CBSE

One cusec is equal to how many liters class 8 maths CBSE

How many crores make 10 million class 7 maths CBSE

Can anyone list 10 advantages and disadvantages of friction