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Grade: 12th pass
        
Can....... someone.... please
 
…solve this..............attachment......... 
one month ago

Answers : (2)

Vikas TU
10366 Points
							
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 
one month ago
Arun
23742 Points
							
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 
one month ago
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