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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 $\left( x,y \right)=\left( a{{t}^{2}},2at \right)$.
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 $2y\dfrac{dy}{dx}=4a$
$\dfrac{dy}{dx}=\dfrac{2a}{y}$
At $\left( x,y \right)=\left[ a{{t}^{2}},2at \right]$
We get, $\Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}$
As $\dfrac{dy}{dx}$ signify the slope of tangent, therefore any tangent on parabola at point

$\left( a{{t}^{2}},2at \right)$ 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, $a{{\left( -1 \right)}^{3}}+\left( -1 \right)\left( 2a-h \right)+k=0$
$\Rightarrow -a-\left( 2a-h \right)+k=0$
$\Rightarrow -3a+h+k=0$
$h+k=3a$
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
hurry,
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.