φ = π – tan –1 2, t = ; φ = π + tan − 1 2, t = ; φ = , t = 0

Sol. The equation of the tangent at (t 2 , 2t) to the parabola y 2 = 4x is

2 ty = 2(x + t 2 )

⇒ ty = x + t 2 ⇒ x – ty + t 2 = 0 ......(1)

The equation of the normal at point on the ellipse 4x 2 + 5y 2 = 20 is

( sec φ ) x – (2 cosec φ ) y = 5 – 4

⇒ ( sec φ ) x – (2 cosec φ ) y – 1 = 0 ........(2)

equation (1) and (2) represents same line

So = = –

t = and t = –

t = cot φ and t = – sin φ

so cot φ = – sin φ ⇒ 4 cos φ = – sin 2 φ

4 cos φ = – (1 – cos 2 φ )

cos 2 φ – 4 cos φ – = 0

⇒ cos 2 φ – 5 cos φ + cos φ – = 0

⇒ cos φ (cos φ – ) + 1 (cos φ – ) = 0

cos φ = or cos φ = –

when φ is in ΙΙ nd quadraint sin φ = , tan φ =– 2 ⇒ φ = π – tan – 1 2

Now t = – sin φ = –

when φ is in ΙΙΙ rd quadrant

sin φ = – , tan φ = 2 ⇒ φ = π + tan – 1 2

and t =

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