Simplify the given equation and graph
x^2-xy+y^2-3=0
(please answer. Needed for my studies tonight. This is about rotation of axes. Thanks
PS: i will really appreciate your response. :)
To simplify the given equation, we start by completing the square.
First, let's rearrange the equation:
x^2 - xy + y^2 - 3 = 0
Next, we want to complete the square for the quadratic terms x^2 - xy + y^2. To do this, we need to find the terms that, when added to both sides of the equation, will create a perfect square trinomial.
To find these terms, we'll take half of the coefficient of the middle term (in this case, -xy), square it, and add it to both sides of the equation. The coefficient of -xy is -1.
Step 1: Take half of -1 and square it: (-1/2)^2 = 1/4.
Step 2: Add 1/4 to both sides of the equation.
x^2 - xy + y^2 - 3 + 1/4 = 1/4.
Now, let's rewrite the left side of the equation as a perfect square trinomial by factoring:
(x^2 - xy + y^2) - 3 + 1/4 = 1/4.
(x - 1/2y)^2 - 3 + 1/4 = 1/4.
Next, simplify the right side of the equation:
(x - 1/2y)^2 - 11/4 = 0.
Now, we have simplified the equation to standard form:
(x - 1/2y)^2 - 11/4 = 0.
To graph this equation, we can plot the points that satisfy the equation. Notice that the equation represents a circle.
The center of the circle is at the point (h, k), where h = 1/2 and k = 0 (since there is no y-term in the perfect square trinomial).
To find the radius of the circle, we take the square root of the constant term on the right side of the equation, which is 11/4. The radius is therefore sqrt(11)/2.
Plot the center at (1/2, 0), and then draw the circle with a radius of sqrt(11)/2 centered at (1/2, 0).
This is how you simplify and graph the given equation: x^2 - xy + y^2 - 3 = 0.