CircleHard
Question
The tangent and normal at P (t), for all real positive t, to the parabola y2 = 4ax meet the axis of the parabola in T and G respectively, then the angle at which the tangent at P to the parabola is inclined to the tangent at P to the circle through the points P, T and G is -
Options
A.cot-1t
B.cot-1t2
C.tan-1t
D.sin-1 
Solution
Equation of tangent and normal at P (at2, 2at) on
y2 = 4ax are
ty = x + at2 ... (1)
y + tx = 2at + at3 ... (2)
So T(-at2, 0) & G (2a + at2, 0)
equation of circle passing P, T & G is
(x + at2) (x-(2a + at2)) + (y - 0) (y - 0) = 0
x2 + y2 - 2ax - at2 (2a + at2) = 0
equation of tangent on the above circle at
P(at2, 2at) is at2x + 2aty - a(x + at2)-at2(2a + at2) = 0
slope of line which is tangent to circle at P
m1 =
slope of tangent at P, m2 = 1/t
∴ tan θ =
⇒ tan θ = t
⇒ θ = tan-1 t = sin-1
y2 = 4ax are
ty = x + at2 ... (1)
y + tx = 2at + at3 ... (2)
So T(-at2, 0) & G (2a + at2, 0)
equation of circle passing P, T & G is
(x + at2) (x-(2a + at2)) + (y - 0) (y - 0) = 0
x2 + y2 - 2ax - at2 (2a + at2) = 0
equation of tangent on the above circle at
P(at2, 2at) is at2x + 2aty - a(x + at2)-at2(2a + at2) = 0
slope of line which is tangent to circle at P
m1 =
slope of tangent at P, m2 = 1/t
∴ tan θ =
⇒ tan θ = t ⇒ θ = tan-1 t = sin-1
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