if the normal at (1,2) on the oarabola y^2=4x meets the parabola again at the point (t^2,2t) then the value of t is
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y² = 4x
differentiate ,
2y dy/dx = 4
dy/dx = 2/y
slope of normal = -1/(dy/dx)
at (1, 2) slope of normal = -1/(2/2) = -1
so, equation of normal
(y -2) = -(x -1)
x + y = 3 -----(1)
question ask eqn (1) touch again parabola , at (t² , 2t)
t² + 2t = 3
t² + 2t -3 = 0
t² + 3t - t -3 = 0
t( t +3 ) -1(t + 3) = 0
(t - 1)(t -3) = 0
t = 1, 3
differentiate ,
2y dy/dx = 4
dy/dx = 2/y
slope of normal = -1/(dy/dx)
at (1, 2) slope of normal = -1/(2/2) = -1
so, equation of normal
(y -2) = -(x -1)
x + y = 3 -----(1)
question ask eqn (1) touch again parabola , at (t² , 2t)
t² + 2t = 3
t² + 2t -3 = 0
t² + 3t - t -3 = 0
t( t +3 ) -1(t + 3) = 0
(t - 1)(t -3) = 0
t = 1, 3
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If the normal at (1,2) on the oarabola y^2=4x meets the parabola again at the point (t^2,2t) then the value of t is
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abhi178
Abhi178 The Sage
y² = 4x
differentiate ,
2y dy/dx = 4
dy/dx = 2/y
slope of normal = -1/(dy/dx)
at (1, 2) slope of normal = -1/(2/2) = -1
so, equation of normal
(y -2) = -(x -1)
x + y = 3 -----(1)
question ask eqn (1) touch again parabola , at (t² , 2t)
t² + 2t = 3
t² + 2t -3 = 0
t² + 3t - t -3 = 0
t( t +3 ) -1(t + 3) = 0
(t - 1)(t +3) = 0
t = 1, -3
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