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Normals are drawn from the point P with slopes m_1 . m_2=\alpha is a part of the parabola itself, then \alpha is equal to
 

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

-2


Answers (1)

best_answer

(b) Let the point P be (h,k).

\begin{array}{ll} \Rightarrow & k=m h-2 m-m^3 \\ \Rightarrow & m^3+m(2-h)+k=0 \\ \Rightarrow & m_1 m_2 m_3=-k \Rightarrow m_3=-\frac{k}{\alpha} \end{array}

\begin{array}{ll} \Rightarrow & \left(-\frac{k}{\alpha}\right)^3-\frac{k}{\alpha}(2-h)+k=0 \\ \\\Rightarrow & k^2=\alpha^2 h-2 \alpha^2+\alpha^3 \end{array}

Thus, the locus of (h,k) is 

\Rightarrow \quad \: \: \: \: \: \: \: \: \: y^2=\alpha^2 x-2 \alpha^2+\alpha^3

On comparing it with y^2=4 x, we get

                                  \alpha ^2=4

and          -2 \alpha^2+\alpha^3=0 \Rightarrow \alpha=2

Posted by

shivangi.shekhar

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