Question

Find the locus of the point N from which 3 normals are drawn to the parabola \[{{y}^{2}}=4ax\] are such that two of them are perpendicular to each other.

Answer 1

Hint: First find the equation of normal which would be a three-degree equation in terms of ‘m’ and put the product of roots -1. By doing this we will get our required solution.

Complete step-by-step answer:
We are given a parabola \[{{y}^{2}}=4ax\] and point N(h, k) from which three normals are drawn. Also, given that two of them are perpendicular that means the product of slopes of two out of three normals is – 1.
That means \[{{m}_{1}}{{m}_{2}}=-1\].
Now, we have to find the locus of the point N(h, k).
We know that any general point on the parabola \[{{y}^{2}}=4ax\] is \[\left( x,y \right)=\left( a{{t}^{2}},2at \right)\].
We know that any line passing from \[\left( {{x}_{1}},{{y}_{1}} \right)\] and the slope m is:
\[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\].
So, the equation of normal at the point \[\left( a{{t}^{2}},2at \right)\] and slope m is:
\[\left( y-2at \right)=m\left( x-a{{t}^{2}} \right)....\left( i \right)\]
Now we take the parabola, \[{{y}^{2}}=4ax\].
Now we differentiate the parabola.
\[\left[ \text{Also, }\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \right]\]
Therefore, we get \[2y\dfrac{dy}{dx}=4a\]
\[\dfrac{dy}{dx}=\dfrac{2a}{y}\]
At \[\left( x,y \right)=\left[ a{{t}^{2}},2at \right]\]
We get, \[\Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}\]
As \[\dfrac{dy}{dx}\] signifies the slope of the tangent, therefore any tangent on parabola at the point \[\left( a{{t}^{2}},2at \right)\] would have a slope \[=\dfrac{1}{t}\].
Now, we know that tangent and normal are perpendicular to each other.
Therefore, \[\left( \text{Slope of tangent} \right)\times \left( \text{Slope of normal} \right)=-1\]
As we have found that \[\text{Slope of tangent =}\dfrac{1}{t}\] and assume that the slope of normal is m. Therefore, we get,
\[\left( \dfrac{1}{t} \right)\times \left( m \right)=-1\].
Hence, \[t=-m\]
Putting the value of t in equation (i), we get,
\[\left[ y-2a\left( -m \right) \right]=m\left[ x-a{{\left( -m \right)}^{2}} \right]\]
\[\Rightarrow y+2am=m\left( x-a{{m}^{2}} \right)\]
\[\Rightarrow y=mx-a{{m}^{3}}-2am\]
By rearranging the given equation, we get,
\[a{{m}^{3}}+m\left( 2a-x \right)+y=0\]
Here, we get a three-degree equation in terms of m, therefore it will have 3 roots \[{{m}_{1}},{{m}_{2}}\] and \[{{m}_{3}}\].
As we know that this normal passes through (h, k), we put x = h and y = k. We get,
\[a{{m}^{3}}+m\left( 2a-h \right)+k=0....\left( ii \right)\]
Comparing the above equation by general three-degree equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\]
We get, \[a=a,\text{ }b=0,\text{ }c=\left( 2a-h \right),\text{ }d=k\]
We know that \[\text{product of roots }=\dfrac{-d}{a}\]
Therefore, we get a product of roots \[=\dfrac{-k}{a}\]
As \[{{m}_{1}},{{m}_{2}}\] and \[{{m}_{3}}\] are roots,
Hence, \[{{m}_{1}}{{m}_{2}}{{m}_{3}}=\dfrac{-k}{a}\]
As we have deduced that \[{{m}_{1}}{{m}_{2}}=-1\],
We get \[\left( -1 \right).{{m}_{3}}=\dfrac{-k}{a}\]
Therefore, \[{{m}_{3}}=\dfrac{k}{a}\]
As \[{{m}_{3}}\] is the root of the equation (ii), therefore, it will satisfy equation (ii).
Now, we put \[{{m}_{3}}\] in equation (ii), we get
\[a{{\left( {{m}_{3}} \right)}^{3}}+{{m}_{3}}\left( 2a-h \right)+k=0\]
As \[{{m}_{3}}=\dfrac{+k}{a}\]
We get, \[a{{\left( \dfrac{k}{a} \right)}^{3}}+\left( \dfrac{k}{a} \right)\left( 2a-h \right)+k=0\]
By simplifying the equation, we get
\[\Rightarrow \dfrac{{{k}^{3}}}{{{a}^{2}}}+\dfrac{k}{a}\left( 2a-h \right)+k=0\]
\[\Rightarrow \dfrac{{{k}^{3}}+ka\left( 2a-h \right)+k{{a}^{2}}}{{{a}^{2}}}=0\]
\[\Rightarrow {{k}^{3}}+ka\left( 2a-h \right)+k{{a}^{2}}=0\]
By taking k common from each term,
We get, \[\left( k \right)\left[ {{k}^{2}}+a\left( 2a-h \right)+{{a}^{2}} \right]=0\]
\[\Rightarrow \left( k \right)\left[ {{k}^{2}}+2{{a}^{2}}-ah+{{a}^{2}} \right]=0\]
Now, to get the locus of N(h, k), we will replace h by x and k by y.
Hence, we get \[\Rightarrow \left( y \right)\left( {{y}^{2}}-ax+3{{a}^{2}} \right)=0\]
\[\Rightarrow {{y}^{2}}-ax+3{{a}^{2}}=0\]
So, the locus of the point N(h, k) is \[\left( {{y}^{2}}-ax+3{{a}^{2}} \right)=0\].

Note: Kindly take care of the negative sign with \[\dfrac{d}{a}\]. Also, many times students confuse with writing the product of roots which is \[-\dfrac{d}{a}\] where d is the constant term of the equation. Also, students need to take care while taking \[{{a}^{2}}\] as LCM.