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From a point \mathrm{(h, k)}  three normals are drawn to the parabola \mathrm{y^2=4 a x} . Now a circle passing through the point of contacts of the normals meet the parabola at four points. Find the equation of circle. 

 

Option: 1

x^2+y^2-(h+2 a) x-\frac{k}{2} y=0


Option: 2

x^2+y^2+(h+2 a) x+\frac{k}{2} y=0


Option: 3

x^2+y^2-(h+a) x+\frac{k}{2} y=0


Option: 4

x^2+y^2-(h+a) x-k y=0


Answers (1)

best_answer

Equation of normal in terms of slope is \mathrm{y=m x-2 a m-a m^3. } Since normals meet at (h, k)

\mathrm{\begin{aligned} & k=m h-2 a m-a m^3 \\ & a m^3+m(2 a-h)+k=0 \\ \therefore \quad & m_1+m_2+m_3=0 \end{aligned} }................(1) & (2)
Let equation of circle be
\mathrm{x^2+y^2+2 g x+2 f y+c=0 }........................(3)
\mathrm{Put x=a^2 \& y=-2 a\, m \, in\, (1)\, then }
\mathrm{\begin{aligned} & \quad a^2 m^4+4 a^2 m^2+2 a g m^2-4 f a m+c=0 \\ & \therefore \quad m_1+m_2+m_3+m_4=0 \end{aligned} }......................(4)
from (2) and (4),
\mathrm{\therefore \quad \mathrm{m}_4=0 }
Hence circle pass through (0,0) i.e., \mathrm{\quad c=0 }

from (1),
\mathrm{\sum \mathrm{m}_1 \mathrm{~m}_2=\mathrm{m}_1 \mathrm{~m}_2+\mathrm{m}_2 \mathrm{~m}_3+\mathrm{m}_3 \mathrm{~m}_1=\frac{(2 \mathrm{a}-\mathrm{h})}{\mathrm{a}} }............(5)
and from (A),
\mathrm{\sum m_1 m_2=\frac{4 a^2+2 a g}{a^2} ) }
\mathrm{\begin{aligned} &\left(\mathrm{m}_1+\mathrm{m}_2\right)\left(\mathrm{m}_3+\mathrm{m}_4\right)+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{m}_3 \mathrm{~m}_4=\frac{4 \mathrm{a}^2+2 \mathrm{ag}}{\mathrm{a}^2} \\ & \Rightarrow \quad\left(\mathrm{m}_1+\mathrm{m}_2\right)\left(\mathrm{m}_3+0\right)+\mathrm{m}_1 \mathrm{~m}_2+0=\frac{4 \mathrm{a}+2 \mathrm{~g}}{\mathrm{a}} \quad\left(\otimes \mathrm{m}_4=0\right) \end{aligned}}
\mathrm{\Rightarrow \quad m_1 m_2+m_2 m_3+m_3 m_1=\frac{4 a+2 g}{a}}..................(6)
from (5) and (6),
\mathrm{\begin{array}{ll} \Rightarrow & 2 \mathrm{a}-\mathrm{h}=4 \mathrm{a}+2 \mathrm{~g} \\ \Rightarrow & 2 \mathrm{~g}=-\mathrm{h}-2 \mathrm{a} \\ & \mathrm{g}=-\frac{(\mathrm{h}+2 \mathrm{a})}{2} \\ \therefore & \end{array}}
\mathrm{and \, \, from (1), }
\mathrm{\mathrm{m}_1 \mathrm{~m}_2 \mathrm{~m}_3=-\frac{\mathrm{k}}{\mathrm{a}} }.....................................................(7)
and from (A), 
\mathrm{\begin{aligned} & \sum \mathrm{m}_1 \mathrm{~m}_2 )\mathrm{~m}_3=\frac{4 \mathrm{fa}}{\mathrm{a}^2} \\ & \mathrm{~m}_1 \mathrm{~m}_2\left(\mathrm{~m}_3+\mathrm{m}_4\right)+\mathrm{m}_3 \mathrm{~m}_4\left(\mathrm{~m}_1+\mathrm{m}_2\right)=\frac{4 \mathrm{f}}{\mathrm{a}} \\ \Rightarrow \quad & \mathrm{m}_1 \mathrm{~m}_2\left(\mathrm{~m}_3+0\right)+0=\frac{4 \mathrm{f}}{\mathrm{a}} \\ & \mathrm{m}_1 \mathrm{~m}_2 \mathrm{~m}_3=\frac{4 \mathrm{f}}{\mathrm{a}} \end{aligned} }\mathrm{\begin{aligned} & \sum \mathrm{m}_1 \mathrm{~m}_2 )\mathrm{~m}_3=\frac{4 \mathrm{fa}}{\mathrm{a}^2} \\ & \mathrm{~m}_1 \mathrm{~m}_2\left(\mathrm{~m}_3+\mathrm{m}_4\right)+\mathrm{m}_3 \mathrm{~m}_4\left(\mathrm{~m}_1+\mathrm{m}_2\right)=\frac{4 \mathrm{f}}{\mathrm{a}} \\ \Rightarrow \quad & \mathrm{m}_1 \mathrm{~m}_2\left(\mathrm{~m}_3+0\right)+0=\frac{4 \mathrm{f}}{\mathrm{a}} \\ & \mathrm{m}_1 \mathrm{~m}_2 \mathrm{~m}_3=\frac{4 \mathrm{f}}{\mathrm{a}} \end{aligned} }...................(8)
from (7) & (8),
\mathrm{\begin{aligned} \frac{4 f}{a} & =-\frac{k}{a} \\ \Rightarrow \quad f & =-\frac{k}{4} \end{aligned} }
Hence equation of circle is
\mathrm{\begin{aligned} & x^2+y^2+2\left\{\frac{-(h+2 a)}{2}\right\} x+2\left\{\frac{-k}{4}\right\} y+0=0 \\ & \Rightarrow \quad x^2+y^2-(h+2 a) x-\frac{k}{2} y=0 \end{aligned} }

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Pankaj Sanodiya

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