A normal, to parabola y2=4 a x whose inclination is 30°, to a parabola cuts it again at an angle of

Question

A normal, to parabola y2=4 a x whose inclination is 30°, to a parabola cuts it again at an angle of (A)  tan -1((√3/2)) (B)  tan -1((2/√3)) (C)  tan -1(2 √3) (D)  tan -1((1/2 √3))

Tardigrade Admin · Accepted Answer

The normal at P(a t12, 2 a t1) is y+x t1=2 a t1+a t13 with slope say tan α=-t1=(1/√3) If it meets curve at Q(a t22, 2 a t2) then t2=-t1-(2/t1)=(7/√3). Then angle θ between parabola (tangent at Q ) and normal at P is given by tan θ=(-t1-(1/t2)/1-(t1)t2)=(1/2 √3) ⇒ θ= tan -1((1/2 √3))

Q. A normal, to parabola $y^{2} = 4 a x$ whose inclination is $3 0^{\circ}$ , to a parabola cuts it again at an angle of

$tan^{- 1} (\frac{3}{2})$

$tan^{- 1} (\frac{2}{3})$

$tan^{- 1} (23)$

$tan^{- 1} (\frac{1}{2 3})$

Q. A normal, to parabola y2=4ax whose inclination is 30∘, to a parabola cuts it again at an angle of

tan−1(23​​)

tan−1(3​2​)

tan−1(23​)

tan−1(23​1​)

Q. A normal, to parabola $y^{2} = 4 a x$ whose inclination is $3 0^{\circ}$ , to a parabola cuts it again at an angle of

$tan^{- 1} (\frac{3}{2})$

$tan^{- 1} (\frac{2}{3})$

$tan^{- 1} (23)$

$tan^{- 1} (\frac{1}{2 3})$