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Question: Find the locus of a point O when three normals are drawn from it are such that one bisects the angle...

Find the locus of a point O when three normals are drawn from it are such that one bisects the angle between the other two.

Explanation

Solution

Hint: Take a general point inside the parabola and draw three normals from that point. Angles between the middle one and the extreme ones should be equal as it bisects them.

Complete step-by-step answer:
Let us first draw a diagram with the details given in the question.


From point PP three normals (PR,PS,PQ)(PR,PS,PQ) are drawn to the parabola y2=4ax{{y}^{2}}=4ax. Equation of a general normal to the parabola y2=4ax{{y}^{2}}=4ax is y=mx−2am−am3y=mx-2am-a{{m}^{3}}, where mm is the slope of the normal.
So we write the equation of the normal as,
y1=mx1−2am−am3{{y}_{1}}=m{{x}_{1}}-2am-a{{m}^{3}}
am3+2am−mx1+y1=0...(i)a{{m}^{3}}+2am-m{{x}_{1}}+{{y}_{1}}=0...(i)
Equation (i)\left( i \right) is a cubic equation in mm with roots m1,m2,m3{{m}_{1}},{{m}_{2}},{{m}_{3}}. It is given that ∠RPS=∠RPQ\angle RPS=\angle RPQ. So we can write,
m3−m21+m3×m2=m2−m11+m2×m1\dfrac{{{m}_{3}}-{{m}_{2}}}{1+{{m}_{3}}\times {{m}_{2}}}=\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{2}}\times {{m}_{1}}}
(m3−m2)(1+m2×m1)=(m2−m1)(1+m3×m2)\left( {{m}_{3}}-{{m}_{2}} \right)\left( 1+{{m}_{2}}\times {{m}_{1}} \right)=\left( {{m}_{2}}-{{m}_{1}} \right)\left( 1+{{m}_{3}}\times {{m}_{2}} \right)
m3(1+m2×m1)−m2(1+m2×m1)=m2(1+m3×m2)−m1(1+m3×m2){{m}_{3}}\left( 1+{{m}_{2}}\times {{m}_{1}} \right)-{{m}_{2}}\left( 1+{{m}_{2}}\times {{m}_{1}} \right)={{m}_{2}}\left( 1+{{m}_{3}}\times {{m}_{2}} \right)-{{m}_{1}}\left( 1+{{m}_{3}}\times {{m}_{2}} \right)
m3+m1×m2×m3−m2−m1(m2)2=m2+m3(m2)2−m1−m1×m2×m3{{m}_{3}}+{{m}_{1}}\times {{m}_{2}}\times {{m}_{3}}-{{m}_{2}}-{{m}_{1}}{{\left( {{m}_{2}} \right)}^{2}}={{m}_{2}}+{{m}_{3}}{{({{m}_{2}})}^{2}}-{{m}_{1}}-{{m}_{1}}\times {{m}_{2}}\times {{m}_{3}}
m3+m1−2m2+2m1×m2×m3−(m2)2(m1+m3)=0.....(ii){{m}_{3}}+{{m}_{1}}-2{{m}_{2}}+2{{m}_{1}}\times {{m}_{2}}\times {{m}_{3}}-{{\left( {{m}_{2}} \right)}^{2}}\left( {{m}_{1}}+{{m}_{3}} \right)=0.....(ii)
Also from equation (i)\left( i \right) we get, the product and sum of roots as,
m1m2m3  = −y1a...(iii){{m}_{1}}{{m}_{2}}{{m}_{3~~}}=~\dfrac{-{{y}_{1}}}{a}...(iii)
m1+m2+m3=0{{m}_{1}}+{{m}_{2}}+{{m}_{3}}=0
−m2= m1+m3...(iv)-{{m}_{2}}=\text{ }{{m}_{1}}+{{m}_{3}}...(iv)
Substituting equations (iii),(iv)\left( iii \right),\left( iv \right) in (ii)\left( ii \right) we get,
−m2−2m2+2(−y1a)−(m2)2(−m2)=0-{{m}_{2}}-2{{m}_{2}}+2\left( \dfrac{-{{y}_{1}}}{a} \right)-{{\left( {{m}_{2}} \right)}^{2}}\left( -{{m}_{2}} \right)=0
(m2)3=3m2+2y1a...(v){{\left( {{m}_{2}} \right)}^{3}}=3{{m}_{2}}+\dfrac{2{{y}_{1}}}{a}...(v)
Since we know that the roots satisfy the equation, we can substitute equation (v)\left( v \right) in (i)\left( i \right),
a(3m2+2y1a)+2am2−(m2×x1)+y1=0a\left( 3{{m}_{2}}+\dfrac{2{{y}_{1}}}{a} \right)+2a{{m}_{2}}-\left( {{m}_{2}}\times {{x}_{1}} \right)+{{y}_{1}}=0
3am2+2y1+2am2−(m2×x1)+y1=03a{{m}_{2}}+2{{y}_{1}}+2a{{m}_{2}}-\left( {{m}_{2}}\times {{x}_{1}} \right)+{{y}_{1}}=0
5am2+3y1−(m2×x1)=05a{{m}_{2}}+3{{y}_{1}}-\left( {{m}_{2}}\times {{x}_{1}} \right)=0
3y1−m2(x1−5a)=03{{y}_{1}}-{{m}_{2}}\left( {{x}_{1}}-5a \right)=0
3y1=m2(x1−5a)3{{y}_{1}}={{m}_{2}}({{x}_{1}}-5a)
m2 =3y1x1−5a...(vi){{m}_{2~}}=\dfrac{3{{y}_{1}}}{{{x}_{1}}-5a}...(vi)
Substituting equation (vi)\left( vi \right) in (v)\left( v \right) we get,

& {{\left( \dfrac{3{{y}_{1}}}{{{x}_{1}}-5a} \right)}^{3}}=3\left( \dfrac{3{{y}_{1}}}{{{x}_{1}}-5a} \right)+\dfrac{2{{y}_{1}}}{a} \\\ & {{\left( \dfrac{3{{y}_{1}}}{{{x}_{1}}-5a} \right)}^{3}}-\left( \dfrac{9{{y}_{1}}}{{{x}_{1}}-5a} \right)-\dfrac{2{{y}_{1}}}{a}=0 \\\ & \dfrac{27{{y}_{1}}^{3}}{{{\left( {{x}_{1}}-5a \right)}^{3}}}-\dfrac{9{{y}_{1}}}{{{x}_{1}}-5a}-\dfrac{2{{y}_{1}}}{a}=0 \\\ \end{aligned}$$ Taking out common terms, we get $$\dfrac{27{{y}_{1}}^{2}}{{{\left( {{x}_{1}}-5a \right)}^{3}}}-\dfrac{9}{{{x}_{1}}-5a}-\dfrac{2}{a}=0$$ Taking the LCM of terms, we get $$\begin{aligned} & \dfrac{27a{{y}_{1}}^{2}-9a{{\left( {{x}_{1}}-5a \right)}^{2}}-2{{\left( {{x}_{1}}-5a \right)}^{3}}}{a{{\left( {{x}_{1}}-5a \right)}^{3}}}=0 \\\ & 27a{{y}_{1}}^{2}-9a{{\left( {{x}_{1}}-5a \right)}^{2}}-2{{\left( {{x}_{1}}-5a \right)}^{3}}=0 \\\ \end{aligned}$$ Using the expansions $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$ and ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$, we get $$\begin{aligned} & 27a{{y}_{1}}^{2}-9a\left( {{x}_{1}}^{2}-10a{{x}_{1}}+25{{a}^{2}} \right)-2\left( {{x}_{1}}^{3}-125{{a}^{3}}-15a{{x}_{1}}\left( {{x}_{1}}-5a \right) \right)=0 \\\ & 27a{{y}_{1}}^{2}-9a\left( {{x}_{1}}^{2}-10a{{x}_{1}}+25{{a}^{2}} \right)-2\left( {{x}_{1}}^{3}-125{{a}^{3}}-15a{{x}_{1}}^{2}+75{{a}^{2}}{{x}_{1}} \right)=0 \\\ & 27a{{y}_{1}}^{2}-9a{{x}_{1}}^{2}+90{{a}^{2}}{{x}_{1}}-225{{a}^{3}}-2{{x}_{1}}^{3}+250{{a}^{3}}+30a{{x}_{1}}^{2}-150{{a}^{2}}{{x}_{1}}=0 \\\ & -2{{x}_{1}}^{3}+21a{{x}_{1}}^{2}-60{{a}^{2}}{{x}_{1}}+25{{a}^{3}}+27a{{y}_{1}}^{2}=0 \\\ \end{aligned}$$ This is the required locus. Now since $$\left( {{x}_{1}},{{y}_{1}} \right)$$ are general points on our locus, we can replace $${{x}_{1}}$$ by $$x$$ and $${{y}_{1}}$$ by $$y$$. So, we will get $$\begin{aligned} & -2{{x}^{3}}+21a{{x}^{2}}-60{{a}^{2}}x+25{{a}^{3}}+27a{{y}^{2}}=0 \\\ & \Rightarrow 2{{x}^{3}}-21a{{x}^{2}}+60{{a}^{2}}x-27a{{y}^{2}}-25{{a}^{3}}=0 \\\ \end{aligned}$$ This is the required locus. Note: Students have to think carefully while deciding which is the variable before eliminating. In this question, students might eliminate $$a$$ which will give them the wrong answer. Also, they may use different techniques to eliminate the variable from the equations. If they feel it is redundant to use $$\left( {{x}_{1}},{{y}_{1}} \right)$$ first and then replace it as $$\left( x,y \right)$$, they may use $$\left( x,y \right)$$ from the start as well.