Grade 12th passAnalytical Geometry
Can....... someone.... please …solve this..............attachment.........
Can....... someone.... please
…solve this..............attachment.........
2 Answers
Vikas TU
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
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
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,
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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