How many integral pairs (x,y) exist such that x² + 10y² - 4xy + 2x + 8y – 18 = 0 ?
(1) two (2) three (3) four (4) six
2006-09-16 01:32:21 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
This can be turned into a quadratic in x. In the standard expression for the x-solutions, the denominator becomes 1, and the expression includes the square root of a quadratic in y. For integral values of x, this quadratic needs to be a perfect square. It can be seen to be a parabola which is positive only between its zeroes, which are just below -3 and just above 1. Its values for -3, -2, -1, 0, and 1 are 1, 19, 25, 19 and 1. Therefore integral y-values of -3, -1 and 1 give two integral x-values each, because of the positive or negative square root. The six integral (x,y) pairs are (-8, -3), (-6, -3), (-8, -1), (2, -1), (0, 1) and (2, 1).
I expect that when you turn it into a quadratic in y, and get a quadratic in x under the square root sign, it has a wider positive range, and more integral square values, but the denominator is 5, so each square value of the x-quadratic can give zero, one, or two integral y values. The final set of answers must be the same, but they need more searching.
2006-09-16 10:51:53 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The equation: x² + 10y² - 4xy + 2x + 8y – 18 = 0 geometrically represents a curve (Conic Section) on XY-plane.
Each pair (x, y) denotes a point on this curve.
So, in fact, there can be infinite number of such points (or pairs) lying on this curve.
2006-09-16 08:45:43 · answer #2 · answered by quidwai 4 · 0⤊ 1⤋
i think its a n eqn of parabola, if so, infinite points r der 2 satisfy this eqn
2006-09-16 08:47:52 · answer #3 · answered by prasadc 2 · 0⤊ 1⤋