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If  \mathrm{A\left(a t_1^2, 2 a t_1\right)} and \mathrm{B\left(t_2^2, 2 a t_2\right)} be the two points on the parabola \mathrm{y^2=4 a x} and \mathrm{A B} cuts the x-axis at C such that \mathrm{A B: A C=3: 1}, then \mathrm{t_2=k t_1} where \mathrm{k=}

Option: 1

1


Option: 2

-1


Option: 3

2


Option: 4

-2


Answers (1)

best_answer

\mathrm{\begin{aligned} & A B: A C=3: 1 \\\\ & \Rightarrow A C: C B=1: 2 \end{aligned}}

\mathrm{\text { Hence } \, \, \, \frac{4 a t_1+2 a t_2}{3}=0 \Rightarrow 2 t_1+t_2=0}

\mathrm{\Rightarrow \mathrm{t}_2=-2 \mathrm{t}_1}

\mathrm{and\, \, \, \, \, \, t_1 t_2=-4 \Rightarrow-2 t_1^2=-4 \, \, \Rightarrow}

\mathrm{t_1^2=-4 \Rightarrow t_1^2=2 \Rightarrow t_1= \pm \sqrt{2}}

Hence coordinates of \mathrm{A \equiv(2 a, \pm 2 \sqrt{2} a) } 

and \mathrm{A \equiv(2 a,-2 \sqrt{2} a), B \equiv\left(8 a,{ }^{\circledR} 4 \sqrt{2} a\right).}

Posted by

Devendra Khairwa

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