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Normals are drawn from the point P with slopes m1, m2, m3 to the parabola x^{2}=4y  . If locus of P with m1m2=\alpha  is a part of parabola itself, then find \alpha .

 

Option: 1

3


Option: 2

5


Option: 3

2

 


Option: 4

6


Answers (1)

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Now, this gives m3 = -k/\alpha  and this must satisfy equation (1).

(-k / \alpha)^3+(-k / \alpha)(2-h)+k=0

Solving this, we get

 k^2=\alpha^2-2 \alpha^2+\alpha^3

y^2=\alpha 2 x-2 \alpha^2+\alpha^3

On comparing this equation with y^2=4x we get 

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

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