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Find the locus of the point \[P\left( h,k \right)\] if three normals drawn from the
 point \[P\] to
\[{{y}^{2}}=4ax\], satisfying the following \[{{m}_{1}}+{{m}_{2}}=1\].

Last updated date: 16th Jul 2024
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Hint: Sum of slopes of three normals of parabola from a particular point is zero.

We are given a parabola \[{{y}^{2}}=4ax\] and point \[P\left( h,k \right)\] from which three
 normals are drawn. Also, given that \[{{m}_{1}}+{{m}_{2}}=1\] that is the sum of slopes of two out of three normals is \[1\].
Now, we have to find the locus of point\[P\left( h,k \right)\].
We know that any general point on parabola \[{{y}^{2}}=4ax\] is \[\left( x,y \right)=\left( a{{t}^{2}},2at
We know that any line passing from \[\left( {{x}_{1}},{{y}_{1}} \right)\] and slope \[m\] is:
\[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\].
So, the equation of normal at 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 \[2y\dfrac{dy}{dx}=4a\]
At \[\left( x,y \right)=\left[ a{{t}^{2}},2at \right]\]
We get, \[\Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}\]
As \[\dfrac{dy}{dx}\] signify the slope of tangent, therefore any tangent on parabola at point

a{{t}^{2}},2at \right)\] would have 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 assumed that slope of normal is \[m\].
Therefore, we get \[\left( \dfrac{1}{t} \right)\times \left( m \right)=-1\].
Hence, \[t=-m\]
Putting value of \[t\]in equation \[\left( i \right)\],
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 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 \[\left( h,k \right)\], we put \[x=h\] and \[y=k\].

We get, \[a{{m}^{3}}+m\left( 2a-h \right)+k=0....\left( ii \right)\]
Comparing 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 \[\text{sum of roots }=\dfrac{-b}{a}\]
As \[b=0\] in equation \[\left( ii \right)\],
Therefore, we get \[\text{sum of roots }=0\].
As \[{{m}_{1}}, {{m}_{2}}\] and \[{{m}_{3}}\] are roots,
Hence, \[{{m}_{1}}+{{m}_{2}}+{{m}_{3}}=0\]
As we have been given that \[{{m}_{1}}+{{m}_{2}}=1\],
We get \[1+{{m}_{3}}=0\]
Therefore, \[{{m}_{3}}=-1\]
As \[{{m}_{3}}\] is root of equation \[\left( ii \right)\],
Therefore, it will satisfy the equation \[\left( ii \right)\].
Now, we put \[{{m}_{3}}\] in equation \[\left( ii \right)\],
We get \[a{{\left( {{m}_{3}} \right)}^{3}}+{{m}_{3}}\left( 2a-h \right)+k=0\]
As, \[{{m}_{3}}=-1\]
We get, \[a{{\left( -1 \right)}^{3}}+\left( -1 \right)\left( 2a-h \right)+k=0\]
\[\Rightarrow -a-\left( 2a-h \right)+k=0\]
\[\Rightarrow -3a+h+k=0\]
Now, to get the locus of \[\left( h,k \right)\], we will replace \[h\] by \[x\] and \[k\] by \[y\].
So, we get \[x+y=3a\]
Hence, we get locus of point \[P\left( h,k \right)\Rightarrow \left( x+y \right)=3a\]

Note: Mistake could be committed in writing the value of sum of roots \[=\dfrac{-b}{a}\] as in
students often write coefficient of second term as \[b\]that is they write \[b=\left( 2a-h \right)\] but
actually \[b\] is the coefficient of \[{{m}^{2}}\] which is zero in the given question.