Transform the quadratic equation to a standard form.
4x^2-y^2-24x-10y+11=0
My answer:
4(x-3)^2 - (y+5)^2 = 0
Therefore, this is a degenerate conic section, right?
4x^2-y^2-24x-10y+11=0
4(x^2-6x) - (y^2+10y) = -11
4(x^2-6x+9) - (y^2+10y+25) = -11 + 4(9) + 25
4(x-3)^2 - (y+5)^2 = 50
Better check to see what you lost while completing those squares.
To transform the given quadratic equation to standard form, you need to complete the square for both the x and y terms. Here's how you can do it step by step:
1. Group the x and y terms separately:
4x^2 - 24x - y^2 - 10y + 11 = 0
2. Complete the square for the x terms:
4(x^2 - 6x) - (y^2 + 10y) + 11 = 0
Take half the coefficient of x (-6) and square it (36). Add and subtract it inside the parentheses:
4(x^2 - 6x + 36 - 36) - (y^2 + 10y) + 11 = 0
Rearrange the terms:
4((x - 3)^2 - 36) - (y^2 + 10y) + 11 = 0
Simplify further:
4(x - 3)^2 - 144 - (y^2 + 10y) + 11 = 0
4(x - 3)^2 - (y^2 + 10y) - 133 = 0
3. Complete the square for the y terms:
4(x - 3)^2 - (y^2 + 10y + 25 - 25) - 133 = 0
Rearrange the terms:
4(x - 3)^2 - (y^2 + 10y + 25) + 25 - 133 = 0
Simplify further:
4(x - 3)^2 - (y + 5)^2 - 108 = 0
Now, the quadratic equation is in standard form:
4(x - 3)^2 - (y + 5)^2 - 108 = 0
From this standard form, we can see that the equation represents a hyperbola, not a degenerate conic section. A degenerate conic section usually has a coefficient of 0 for either the x^2 or y^2 term, but in this case, both coefficients are nonzero.