Question bank/JEE Main/Mathematics
JEE MainMathematicsModerateConic Sections
P is a point on the circle x^2 + y^2 = 1. Line OP, where O is
origin, and x = 1 meet at Q. L_1 is a line parallel to x-axis drawn
from Q. A line is drawn parallel to x axis from P meeting x = 1 at R. OR
meets L_1 at S. Then locus of S is
Show answer and solution
[IMAGE] Let point P be (cosθ, sinθ), so equation of OP is y = (tanθ)x, hence
point Q is (1, tanθ). Equation of L_1 is y = tanθ. Now equation of
line PR is y = sinθ, hence point R is (1, sinθ). Therefore equation of
OR is y = (sinθ)x. Point of intersection of OR and L_1 is S(secθ,
tanθ). Hence locus of S is x^2 - y^2 = 1, a hyperbola
ekuk fcUnq P(cosθ, sinθ) gS blhfy, OP dk lehdj.k y = (tanθ)x gSA vr sinθ gSA vr ++ fcUnq R(1, sinθ) gS blhfy, OR dk lehdj.k y = (sinθ)x
fcUnqiFk x^2 - y^2 = 1
point Q is (1, tanθ). Equation of L_1 is y = tanθ. Now equation of
line PR is y = sinθ, hence point R is (1, sinθ). Therefore equation of
OR is y = (sinθ)x. Point of intersection of OR and L_1 is S(secθ,
tanθ). Hence locus of S is x^2 - y^2 = 1, a hyperbola
ekuk fcUnq P(cosθ, sinθ) gS blhfy, OP dk lehdj.k y = (tanθ)x gSA vr sinθ gSA vr ++ fcUnq R(1, sinθ) gS blhfy, OR dk lehdj.k y = (sinθ)x
fcUnqiFk x^2 - y^2 = 1
Source: Grass-G / M-27 Mathematics Paper 2 15 Jan.docx
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