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Let A and B be two points on a parabola $\mathrm{y^2=x}$ with vertex V such that VA is perpendicular to VB and $\mathrm{ \theta}$ is the angle between the chord VA and the axis of the parabola. The value of $\mathrm{\frac{|V A|}{|V B|}}$ is
Option: 1
$\tan \theta$

Option: 2
$\tan ^3 \theta$

Option: 3
$\cot ^2 \theta$

Option: 4
$\cot ^3 \theta$

Answers (1)

best_answer

$\mathrm{\tan \theta=\frac{2 a t_1}{a t_1^2}=\frac{2}{t_1} \quad\left[\text { Here, } a=\frac{1}{4}\right]}$

Since, VA ⊥ VB

$\mathrm{\begin{aligned} & \Rightarrow m_1 \times m_2=-1 \\ & \Rightarrow \frac{2}{t_1} \times \frac{2}{t_2}=-1 \\ & \Rightarrow t_1 t_2=-4 \end{aligned}}$

$\mathrm{\text { Now, }|V A|=\sqrt{a^2 t_1^4+4 a^2 t_1^2}}$

$\mathrm{\left.=a t_1 \sqrt{t_1^2+4} \quad \text { (assuming } t_1>0\right)}$

$\mathrm{\begin{aligned} & \text { and }|V B|=\sqrt{a^2 t_2^4+4 a^2 t_2^2}=a\left|t_2\right| \sqrt{t_2^2+4} \\ & =a\left|\frac{-4}{t_1}\right| \sqrt{\left(\frac{-4}{t_1}\right)^2+4}=\frac{8 a}{t_1^2} \sqrt{4+t_1^2} \\ & \text { Thus, } \frac{|V A|}{|V B|}=\frac{t_1^3}{8} \end{aligned}}$

$\mathrm{\text { Also, we have, } \tan \theta=\frac{2}{t_1}}$

$\mathrm{\Rightarrow \frac{t_1}{2}=\cot \theta \Rightarrow \frac{t_1^3}{8}=\cot ^3 \theta}$

$\mathrm{\text { Hence, } \frac{|V A|}{|V B|}=\cot ^3 \theta}$

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