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Can....... someone.... please …solve this..............attachment.........

Can....... someone.... please
 
…solve this..............attachment......... 

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Grade:12th pass

2 Answers

Vikas TU
14149 Points
3 years ago
Let point on parabola be P(at^2, 2at) and S(a, 0) also SPM is equilateral triangle M is foot of perpendicular from point on parabola to directrix ∴ M is (– a, 2at) Now SPM is equilateral triangle
SP = PM = SM 
SM^2 = PM^2
(a+a)^2 + (-2at)^2 = (at^2+a)^2 
t^4 -2t^2 -3 = 0 
t = sqrt(3)
SP = 4a, 
Hope this will help 
Good Luck 
Arun
25757 Points
3 years ago
SP = PM = SM 
SM^2 = PM^2
(a+a)^2 + (-2at)^2 = (at^2+a)^2 
t^4 -2t^2 -3 = 0 
t = sqrt(3)
SP = 4a, 
 
Hence option C is correct 

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