Question

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 ${{m}_{1}}+{{m}_{2}}=1$.

Answer 1

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

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.