Question bank/JEE Main/Mathematics
JEE MainMathematicsToughCircles
Vertices of a variable triangle are (3, 4), (5cosθ, 5sinθ) and
(5sinθ, - 5cosθ). Then locus of its orthocenter is,d pj f=Hkqt (3, 4), (5cosθ, 5sinθ) rFkk (5sinθ, - 5cosθ)
gSA rc f=Hkqt
Show answer and solution
OA = OB = OC (0, 0) is circumcentre ifjdsUæ gS
as pawfd OG: GH = 1: 2 H = 3G
x = 3 + 5cosθ + 5sinθ and vkSj y = 4 + 5sinθ - 5cosθ
x + y = 7 + 10sinθ and vkSj x - y = - 1 + 10cosθ
(x + y - 7)^2 + (x - y + 1)^2 = 10^2
as pawfd OG: GH = 1: 2 H = 3G
x = 3 + 5cosθ + 5sinθ and vkSj y = 4 + 5sinθ - 5cosθ
x + y = 7 + 10sinθ and vkSj x - y = - 1 + 10cosθ
(x + y - 7)^2 + (x - y + 1)^2 = 10^2
Source: Grass-G / M-27 Mathematics Paper 2 15 Jan.docx
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