qudratic

a) The equation required is x^2 - y - 4 + k(x - y) = 0

b)Put x=5, y=7 into the equation found in (a),
5^2 - 7 - 4 + k(5 - 7) = 0
14 - 2k = 0
k = 7

Hence, the equation required is x^2 - y - 4 + 7(x - y) = 0
i.e. x^2 + 7x -8y - 4 = 0
(or y = 1/8 x^2 + 7/8 x - 1/2)

2007-07-21 09:27:06 補充：
x^2 - 4 - y k(x - y) = 0 ...(1)This notation represents a family of parabola passing through two fixed points.Let the two lines x^2 - 4 -y =0 and x - y = 0 intersect at a point (a, b).Then (a, b) satisfies both equation which give 0.Substitute (a, b) into (1), LHS = 0 k(0) = 0.

2007-07-21 09:27:19 補充：
Therefore, for any value of k (not equal to -1), (1) can find the parabola passing through the two fixed points.For k = -1, there is no y in the equation. No graphs can be drawn.So it is better to state clearly that k =/= -1 at first.

a)
x^2 - y - 4 + k(x - y) = 0, where k is constant ------------ (1)

b)
sub (5,7) into (1),
25 - 7 - 4 + k(5 - 7) = 0
k = 7

sub k = 7 back into (1),
x^2 - y - 4 + (7)(x - y) = 0
x^2 + 7x - 4 = 8y