A circle intersects a parabola y=x^2 at A(1,1), B (3,9) C(-6,36) and D(p,q) find point D

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A circle intersects a parabola y=x^2 at A(1,1), B (3,9) C(-6,36) and D(p,q) find point D

[From: ] [author: ] [Date: 12-01-01] [Hit: ]

k = 13, r = sqrt(1105)(x + 30)^2 + (y - 13)^2 = 1105y = x^2solve simultaneously:D(2 ,......

(x + 30)^2 + (x^2 - 13)^2 = 1105
x^2 + 60x + 900 + x^4 - 26x^2 + 169 = 1105
x^4 - 25x^2 + 60x - 36 = 0

Ugh, quadratics are ugly, but we know 3 roots are 1, 3, -6, so call the left side above p(x) and divide this by q(x) = (x - 1)(x - 3)(x + 6) to get the final root. The divisor is:

q(x) = (x^2 - 4x + 3)(x + 6) = x^3 + 2x^2 - 21x - 18

p(x) - x*q(x) = (x^4 - 25x^2 + 60x - 36) - (x^4 + 2x^3 - 21x^2 - 18x)
= -2x^3 - 4x^2 + 78x - 36 = -2*q(x)
p(x) - x*q(x) = -2 q(x)
p(x) = (x - 2)q(x)

So, the final root of p(x) is x=2, with y = x^2 = 4, giving (2,4) as the final point.

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Three points determine a circle.
Let the equation of the circle be (x-a)^2 + (y-b)^2 = r^2
Plug in the three given points,
(1-a)^2 + (1-b)^2 = r^2
(3-a)^2 + (9-b)^2 = r^2
(-6-a)^2 + (36-b)^2 = r^2
Once you get a, b and r, plug in D(p, p^2) to find point D.

Can you finish it?

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assuming the circle's eqn is (x-a)^2 + (y-b)^2 = c^2
subsititute A B C
(1-a)^2 + (1-b)^2 = c^2 --1)
(3-a)^2 +(9-b)^2 = c^2 --2)
(-6-a)^2 + (36-b)^2 = c^2--3)

solve these three eqns, you get a b c
then you can get the intersection between this circle and the parabola

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(x - h)^2 + (y - k)^2 = r^2
h^2 + k^2 = r^2
(x - h)^2 + (x^2 - k)^2 = r^2
x = 1=> h^2 -2h + k^2 - 2k - r^2 = -2
x = 3 => h^2 - 6h + k^2 - 18k - r^2 + 90 = 0
x = -6 => h^2 + 12h + k^2 - 72k - r^2 + 1332 = 0
h = -30, k = 13, r = sqrt(1105)
(x + 30)^2 + (y - 13)^2 = 1105
y = x^2
solve simultaneously:
D(2 , 4)

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