If Two tangents drawn through the point $(α, β)$ to the parabola $y^{2} = 4 x$ such that slope of one tangent is 3 times of the other then

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$β = \frac{2}{9} α^{2}$

$α = \frac{3}{16} β^{2}$

$2 α = 9 β^{2}$

$2 β = 9 α^{2}$

answer is B.

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Detailed Solution

Given parabola $y^{2} = 4 x \Rightarrow a = 1$

Given point $(x_{1}, y_{1}) = (α, β) a n d m_{1} = 3 m_{2}$

$m_{1} m_{2} = \frac{β}{α} S 11 = β^{2} - 4 α \frac{{m_{1}}^{2}}{3} = \frac{β}{α} \tan θ = L e t y = m x + \frac{1}{m} b e \tan g e n t t o g i v e n p a r a b o l a S i n c e i t p a s s e s t h r o u g h t h e p o i n t (α, β) t h e n \Rightarrow β = α m + \frac{1}{m} \Rightarrow α m^{2} - β m + 1 = 0 l e t m_{1,} m_{2} b e r o o t s w h i c h a r e s l o p e s o f \tan g e n t s w h e r e m_{1} + m_{2} = \frac{β}{α} \to (1) a n d m_{1} m_{2} = \frac{1}{α} \to (2) G i v e n m_{1} = 3 m_{2} (1) b e c o m e s 4 m_{2} = \frac{β}{α} \Rightarrow m_{2} = \frac{β}{4 α} (2) b e c o m e s 3 {m^{2}}_{2} = \frac{1}{α} \Rightarrow 3 {(\frac{β}{4 α})}^{2} = \frac{1}{α} \Rightarrow \frac{3 β^{2}}{16 α^{2}} = \frac{1}{α} \Rightarrow 3 β^{2} = 16 α \Rightarrow α = \frac{3}{16} β^{2}$