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Another Algebra Problem

Find the x and y values of the intersection points of the following two equations:

1) ax + by + c =0;
2) x^2 + y^2 + dx + ey + f = 0.

Good Luck!

766 day(s) ago

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Answers (2)

jcatta21 		The two equations are both algebraic elementary structures with a linear and conic equation. The first is the general form of an algebraic linear polynomial and the second is a general conic structure defining formally as Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0. In the second equation, the coefficient of Bxy is zero yielding an excluded binomial term, which would be required for graphing linear transformations. In order to find the x and y values of the intersection points of the two algebraic equations, there could be one of several approaches such as completing the square. By using the completing the square method, you would have to arrange the second equation in standard to reflect the form a(x +- h)^2 +-k. Thus, the x and y values of the intersection points of the two equations can be derived as follows:

x^2 + y^2 + dx + ey + f = 0 => x^2 + dx + y^2 + ey + f = 0 => x^2 + dx + y^2 + ey = -f => 1(x^2 + dx) + 1(y^2 + ey) = -f. When completing the square, the standard form is derived after taking one-half of the binomial or first-degree term. Thus, (1/2*b)^2 = (d/2)^2 = d^2/4 => 1(x^2 + dx + d^2/4) + 1(y^2 + dy + d^2 /4) = -f + d^2/2 thus yielding perfect squares for both x and y terms as (x + d/2)^2 + (y + d/2)^2 = d^2 - 2f/2. By dividing out the constant (d^2 - 2f/2) on both sides yields the standard equation of a conic as follows:

(x + d/2)^2/2/(d^2 - 2f) + (y + d/2)^2 / 2/(d^2 - 2f) = 1 => (x - h)^2 / a^2 +- (y - k)^2 / b^2 = 1. Given the standard equation of a conic, both x and y terms can be derived easily. The type of conic given is seemingly an Ellipse with the sum of the square between the (x, y) values equal to a foci c. With this formula defined, the x and y terms can be derived through the distance formula c^2 = a^2 - b^2.

J.C


Posted 765 dy ago

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matherapist 		I don't know of a real simple way to do this, but eqn 2 is a circle of the form

(x+d/2)^2 + (y+e/2)^2 = -f + (d^2)/4 + (e^2)/4 This is done by collecting terms of the same variable and completing the square.

Eqn 1 can be solved for x or y and plugged into the above and then solved for the appropriate variable.

This is too much typing and effort for me.

Another view is to subtract eqn 1 from eqn2 and get

x^2 +y^2 +(d-a)x +( e-b)y +f-c =0

This can be put in the form of

[x+(d-a)/2]^2 + [y+(e-b)/2]^2= f-c+.25(d-a)^2+.25(e-b)^2

which is a circle centered at -(d-a)/2, (e-b)/2 of radius f-c+.25(d-a)^2+.25(e-b)^2 which means x and y lies on the circle and the 2 equations do not intersect




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