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Three normals with slopes \mathrm{m_1, m_2\: and \: m_3} are drawn from a point \mathrm{P} not on the axis of the parabola \mathrm{y^2=4 x.\: If\: m_1 m_2=\alpha}, results in the locus of \mathrm{P} being a part of the parabola, find the value of \mathrm{\alpha }
 

Option: 1

4


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

Any normal of slope \mathrm{m} to the parabola \mathrm{y^2=4 x} is

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

If it passes through \mathrm{ (h, k) }, then

\mathrm{ k=m h-2 m-m^3 }

\mathrm{ \Rightarrow m^3+(2-h) m+k=0 }

Thus \mathrm{ m_1 m_2 m_3=-k }
Now \mathrm{ m_1 m_2=\alpha (given) \Rightarrow m_3=-\frac{k}{\alpha} }

Now \mathrm{ m_3 } satisfies equation (2)

\mathrm{so -\frac{k^3}{\alpha^3}-(2-h) \frac{k}{\alpha}+k=0}

\mathrm{\Rightarrow k^3+(2-h) k \alpha^2-k \alpha^3=0}

Thus locus of \mathrm{P} is

\mathrm{y^3+(2-x) y \alpha^2-y \alpha^3=0 }

\mathrm{\Rightarrow y^2+(2-x) \alpha^2-\alpha^3=0, \text { as } y \neq 0 }

( \mathrm{P } does not lie on the axis of the parabola)

\mathrm{ \Rightarrow y^2=\alpha^2 x-2 \alpha^2+\alpha^3 }

If it is a part of the parabola \mathrm{y^2=4 x}, then

\mathrm{ \begin{aligned} & \alpha^2=4 \text { and }-2 \alpha^2+\alpha^3=0 \\ \Rightarrow & \alpha=2 \end{aligned} }

Hence option 3 is correct.
 

Posted by

jitender.kumar

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