Let $A$ and $B$ be two points on a parabola $y^{2} = x$ with vertex $V$ such that $VA$ is perpendicu

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

Accepted Answer

Sol. [Key Concept: Parametrise $A,B$ by the chord's inclination $	heta$ from the vertex; $VA$ perpendicular to $VB$ and the distances give $\frac{|VA|}{|VB|} = \cot^{3}	heta$.]$${VA\bot VB \Rightarrow \frac{2}{t_{1}}\frac{2}{t_{2}} = - 1 \Rightarrow t_{1}t_{2} = - 4 }{\left| \frac{VA}{VB} ight| = \frac{\sqrt{\frac{1}{4}t_{1}^{2} + \frac{1}{16}t_{1}^{4}}}{\sqrt{\frac{1}{4}t_{2}^{2} + \frac{1}{16}t_{2}^{4}}} = \frac{\left| t_{1} ight|\sqrt{4 + t_{1}^{2}}}{\left| t_{2} ight|\sqrt{4 + t_{2}^{2}}} }{\frac{\left| t_{1} ight|\sqrt{4 + t_{1}^{2}}}{\left| - \frac{4}{t_{1}} ight|\sqrt{4 + \frac{16}{t_{1}^{2}}}} = \frac{\left| t_{1}^{3} ight|}{8} = \frac{8\cot^{3}	heta}{8} = \cot^{3}	heta }{\left\lbrack \because tan	heta = \frac{2}{t_{1}} ightbrack }$$

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