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Three normals from a point to the parabola \mathrm{y^2=4 a x} meet the axis of the parabola in points whose abscissa are in A.P. Find the locus of the point.

 

Option: 1

\mathrm{9 a y^2=2(x-2 a)^3}


Option: 2

\mathrm{27 a y^2=2(x-2 a)^2}


Option: 3

\mathrm{2 a y^2=27(x-2 a)^3}


Option: 4

\mathrm{27 a y^2=2(x-2 a)^3}


Answers (1)

best_answer

The equation of any normal to the parabola is 

\mathrm{y=m x-2 a m-a m^3}

If it passes through the point (h, k) then

\mathrm{a m^3+m(2 a-h)+k=0} …(i)

The normal cuts the axis of the parabola viz., y = 0 at point where \mathrm{x=2 a+a m^2}

Hence the abscissa of the points in which the normals through (h, k) meet the axis of the parabola are 

\mathrm{x_1=2 a+a m_1^2, x_2=2 a+a m_2^2, x_3=2 a+a m_3^2}

\mathrm{\text { Since } x_1, x_2, x_3 \text { are in A.P. }}

\mathrm{\begin{aligned} & \left(2 a+a m_1^2\right)+\left(2 a+a m_3^2\right)=2\left(2 a+a m_2^2\right) \\ \Rightarrow \quad & m_1^2+m_3^2=2 m_2^2 \end{aligned}}    …(ii)

\mathrm{\text { Also, from (i), } m_1+m_2+m_3=0}                                 …(iii)

\mathrm{m_2 m_3+m_3 m_1+m_1 m_2=\frac{2 a-h}{a}}                                  …(iv)

\mathrm{\text { and } m_1 m_2 m_3=-\frac{k}{a}}                                                              …(v)

From (iii),

\mathrm{\begin{aligned} & \left(m_1+m_3\right)^2=m_2^2 \Rightarrow m_1^2+m_3^2+2 m_1 m_3=m_2^2 \\ \Rightarrow & 2 m_2^2-2 \cdot \frac{k}{a m_2}=m_2^2 \\ \Rightarrow & m_2^2=\frac{2 k}{a m_2} \\ \Rightarrow \quad & a m_2^3=2 k \end{aligned}}     …(vi)

\mathrm{\text { Since } m_2 \text { is a root of (i), } \quad a m_2^3+m_2(2 a-h)+k=0}

\mathrm{\begin{aligned} & \Rightarrow \quad 2 k+m_2(2 a-h)+k=0 \\ & \Rightarrow \quad\left\{m_2(h-2 a)\right\}^3=27 k^3 \\ & \Rightarrow \quad \frac{2 k}{a}(h-2 a)^3=27 k^3 \\ & \Rightarrow \quad 27 a k^2=2(h-2 a)^3 \end{aligned}}

\mathrm{\text { Hence the locus of }(h, k) \text { is } 27 a y^2=2(x-2 a)^3 \text {. }}

Posted by

vishal kumar

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