. Prove that, the normal to


y2 =12x at (3, 6) meets the parabola again in (27, −18) & circle on this normal chord as diameter is x2 + y2 − 30x + 12y – 27 = 0

Best Answer

y2 =12x = 4 × 3 × x

a = 3

point (3, 6) compare it with (at2, 2at)

we get 6=2at

t = 6/2a    ​=6/2 × 3 ​= 1

equation of normal

y + Tx = 2at + at3

y + x = 6 + 3

y = 9 − x

on this normal intersects again as parabola

y2 = 12x

or (9−x)2 = 12x

or 81−18x + x2 = 12x

x2−30x + 81 = 0

x2 − 27x − 3x + 81=0

x(x−27) −3(x−27) =0

(x−3) (x−27) =0

x = 3 or 27

x =3 is already taken as is point of normal

x=27 then y=9 – x = 9 −27=−18

Hence parabola meets normal at (27, −18)

for required circle

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