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Hint- Apply the concept of co-normal points over here. Co normal points are the feet of any three normal that are drawn from any point. Note that, there can be at the most, only 3 normals possible to be drawn on a parabola from a single point. Youâ€™ll eventually get a cubic equation in their slopes, and apply general results of cubic equations, like sum of roots, sum of product of roots taken two at a time, and the product of the roots, to get to the final answer.

Letâ€™s assume a parabola \[{{x}^{2}}=8y\].

General equation of normal of the parabola \[{{x}^{2}}=8y\]is

\[x=ym-2bm-b{{m}^{3}}\] â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

Where \[\dfrac{1}{m}=\]slope of normal

As we can see that the equation of normal \[x=ym-2bm-b{{m}^{3}}\]is a $3$ degree polynomial in $m$.

Therefore; this equation will have three solutions.

It means three normals can be drawn on a parabola from one point, lying anywhere.

Let the point of the intersection of normal is \[C\left( h,k \right)\].

Therefore, point $C$ will satisfy the equation (1).

From equation (1) and the point $C$, we get :

\[h=km-2bm-b{{m}^{3}}\]

\[\Rightarrow b{{m}^{3}}+m\left( 2b-k \right)+h=0\] â€¦â€¦â€¦.. (2)

Let \[{{m}_{1}},{{m}_{2}}\,and\,{{m}_{3}}\]are solutions of equation (2).

Therefore,

\[{{m}_{1}}+{{m}_{2}}+{{m}_{3}}=0\],

\[{{m}_{1}}{{m}_{2}}+{{m}_{2}}{{m}_{3}}+{{m}_{3}}{{m}_{1}}=\dfrac{\left( 2b-k \right)}{b}\] and

\[{{m}_{1}}{{m}_{2}}{{m}_{3}}=-\dfrac{h}{b}\] â€¦â€¦.. (3)

Letâ€™s assume \[{{m}_{1}}\,and\,{{m}_{2}}\,\]are the slopes of two normals which intersect normally each other at point $C$.

Therefore, the product of their slopes \[=-1\]

\[{{m}_{1}}{{m}_{2}}=-1\] â€¦â€¦â€¦. (4)

Put the value of \[{{m}_{1}}{{m}_{2}}=-1\]in equation (3)

Therefore, from equation (3) and (4), we get :

\[\left( -1 \right){{m}_{3}}=-\dfrac{h}{b}\]

\[\Rightarrow {{m}_{3}}=\dfrac{h}{b}\] â€¦â€¦â€¦ (5)

As \[{{m}_{3}}\]is also a solution to equation (2), it will satisfy the equation.

Therefore, from equation (2) and (5)

Put the value of \[{{m}_{3}}=\dfrac{h}{b}\] in equation (3), we get :

\[\Rightarrow b{{\left( \dfrac{h}{b} \right)}^{3}}+\dfrac{h}{b}\left( 2b-k \right)+h=0\]

\[\Rightarrow {{\dfrac{h}{{{b}^{2}}}}^{3}}+2h-\dfrac{hk}{b}+h=0\]

Taking $h$ common, we get :

\[\Rightarrow h\left( \dfrac{{{h}^{2}}}{{{b}^{2}}}+3-\dfrac{k}{b} \right)=0\]

\[\therefore \dfrac{{{h}^{2}}}{{{b}^{2}}}+3-\dfrac{k}{b}=0\]

\[\Rightarrow {{h}^{2}}+3{{b}^{2}}-kb=0\]

\[\Rightarrow {{h}^{2}}=b\left( k-3b \right)\] â€¦â€¦â€¦.. (6)

Interchange \[\left( h,k \right)\to \left( x,y \right)\]and equation (6) becomes

\[\Rightarrow {{x}^{2}}=b\left( y-3b \right)\] â€¦â€¦â€¦â€¦. (7)

According to the question, the given parabola is \[{{x}^{2}}=8y\].

Comparing with the general equation of a parabola \[{{x}^{2}}=4ay\]

\[{{x}^{2}}=4by\] â€¦â€¦â€¦ (A)

\[{{x}^{2}}=8y\] â€¦â€¦â€¦â€¦ (B)

From (A) and (B),

\[b=2\] Put this value in equation (7)

Form equation (7)

\[\Rightarrow {{x}^{2}}=2\left( y-3\times 2 \right)\]

\[\Rightarrow {{x}^{2}}=2\left( y-6 \right)\] Locus of point of intersection.

Note: We can also start from parabola \[{{y}^{2}}=4ax\]and there general equation of normal\[y=xm-2am-a{{m}^{3}}\]. But at the end of calculation just interchange the values

\[x\to y,y\to x\,and\,a\to b\].

Letâ€™s assume a parabola \[{{x}^{2}}=8y\].

General equation of normal of the parabola \[{{x}^{2}}=8y\]is

\[x=ym-2bm-b{{m}^{3}}\] â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

Where \[\dfrac{1}{m}=\]slope of normal

As we can see that the equation of normal \[x=ym-2bm-b{{m}^{3}}\]is a $3$ degree polynomial in $m$.

Therefore; this equation will have three solutions.

It means three normals can be drawn on a parabola from one point, lying anywhere.

Let the point of the intersection of normal is \[C\left( h,k \right)\].

Therefore, point $C$ will satisfy the equation (1).

From equation (1) and the point $C$, we get :

\[h=km-2bm-b{{m}^{3}}\]

\[\Rightarrow b{{m}^{3}}+m\left( 2b-k \right)+h=0\] â€¦â€¦â€¦.. (2)

Let \[{{m}_{1}},{{m}_{2}}\,and\,{{m}_{3}}\]are solutions of equation (2).

Therefore,

\[{{m}_{1}}+{{m}_{2}}+{{m}_{3}}=0\],

\[{{m}_{1}}{{m}_{2}}+{{m}_{2}}{{m}_{3}}+{{m}_{3}}{{m}_{1}}=\dfrac{\left( 2b-k \right)}{b}\] and

\[{{m}_{1}}{{m}_{2}}{{m}_{3}}=-\dfrac{h}{b}\] â€¦â€¦.. (3)

Letâ€™s assume \[{{m}_{1}}\,and\,{{m}_{2}}\,\]are the slopes of two normals which intersect normally each other at point $C$.

Therefore, the product of their slopes \[=-1\]

\[{{m}_{1}}{{m}_{2}}=-1\] â€¦â€¦â€¦. (4)

Put the value of \[{{m}_{1}}{{m}_{2}}=-1\]in equation (3)

Therefore, from equation (3) and (4), we get :

\[\left( -1 \right){{m}_{3}}=-\dfrac{h}{b}\]

\[\Rightarrow {{m}_{3}}=\dfrac{h}{b}\] â€¦â€¦â€¦ (5)

As \[{{m}_{3}}\]is also a solution to equation (2), it will satisfy the equation.

Therefore, from equation (2) and (5)

Put the value of \[{{m}_{3}}=\dfrac{h}{b}\] in equation (3), we get :

\[\Rightarrow b{{\left( \dfrac{h}{b} \right)}^{3}}+\dfrac{h}{b}\left( 2b-k \right)+h=0\]

\[\Rightarrow {{\dfrac{h}{{{b}^{2}}}}^{3}}+2h-\dfrac{hk}{b}+h=0\]

Taking $h$ common, we get :

\[\Rightarrow h\left( \dfrac{{{h}^{2}}}{{{b}^{2}}}+3-\dfrac{k}{b} \right)=0\]

\[\therefore \dfrac{{{h}^{2}}}{{{b}^{2}}}+3-\dfrac{k}{b}=0\]

\[\Rightarrow {{h}^{2}}+3{{b}^{2}}-kb=0\]

\[\Rightarrow {{h}^{2}}=b\left( k-3b \right)\] â€¦â€¦â€¦.. (6)

Interchange \[\left( h,k \right)\to \left( x,y \right)\]and equation (6) becomes

\[\Rightarrow {{x}^{2}}=b\left( y-3b \right)\] â€¦â€¦â€¦â€¦. (7)

According to the question, the given parabola is \[{{x}^{2}}=8y\].

Comparing with the general equation of a parabola \[{{x}^{2}}=4ay\]

\[{{x}^{2}}=4by\] â€¦â€¦â€¦ (A)

\[{{x}^{2}}=8y\] â€¦â€¦â€¦â€¦ (B)

From (A) and (B),

\[b=2\] Put this value in equation (7)

Form equation (7)

\[\Rightarrow {{x}^{2}}=2\left( y-3\times 2 \right)\]

\[\Rightarrow {{x}^{2}}=2\left( y-6 \right)\] Locus of point of intersection.

Note: We can also start from parabola \[{{y}^{2}}=4ax\]and there general equation of normal\[y=xm-2am-a{{m}^{3}}\]. But at the end of calculation just interchange the values

\[x\to y,y\to x\,and\,a\to b\].

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