If by x<2 you mean x^2 (x-squared) then
(a) still garbled
(b) x^2 - 4xy + 3x + 25y - 6 = 0
(x + 3/2)^2 - 4xy + 25y = 33/4
is an hyperbola rotated through
tan2θ = -4
θ ≈ 52°
a) x<2 -4xy + 10 <2+2y-5 = 0
b) x<2 -4xy + 3x + 25y -6= 0
(a) still garbled
(b) x^2 - 4xy + 3x + 25y - 6 = 0
(x + 3/2)^2 - 4xy + 25y = 33/4
is an hyperbola rotated through
tan2θ = -4
θ ≈ 52°
To determine the nature of the conic, we can use the discriminant. The discriminant of a conic is given by the formula:
D = B^2 - 4AC
Where A, B, and C are the coefficients of x^2, xy, and y^2 respectively.
In this case:
A = 1
B = -4
C = 0
Substituting these values into the discriminant formula, we get:
D = (-4)^2 - 4(1)(0)
= 16
Since the discriminant is positive (D > 0), the conic is an ellipse.
b) The equation given is x^2 - 4xy + 3x + 25y - 6 = 0.
Again, we can use the discriminant to determine the nature of the conic.
A = 1
B = -4
C = 0
D = (-4)^2 - 4(1)(0)
= 16
The discriminant is positive, so the conic is an ellipse.
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
Now, let's analyze each equation individually:
a) x^2 - 4xy + 10x + 2y - 5 = 0
Comparing this equation with the general form, we can see that:
- A = 1 (coefficient of x^2)
- B = -4 (coefficient of xy)
- C = 0 (since there is no y^2 term)
- D = 10 (coefficient of x)
- E = 2 (coefficient of y)
- F = -5 (constant term)
To determine the nature of this conic, we can calculate the discriminant, which is given by the formula:
D = B^2 - 4AC
Substituting the values from our equation:
D = (-4)^2 - 4(1)(0) = 16 - 0 = 16
If the discriminant (D) is positive and non-zero, the conic is an ellipse.
If the discriminant (D) is zero, the conic is a parabola.
If the discriminant (D) is negative, the conic is a hyperbola.
In this case, since D = 16 (positive and non-zero), the conic given by equation (a) is an ellipse.
b) x^2 - 4xy + 3x + 25y - 6 = 0
Comparing this equation with the general form, we can see that:
- A = 1 (coefficient of x^2)
- B = -4 (coefficient of xy)
- C = 0 (since there is no y^2 term)
- D = 3 (coefficient of x)
- E = 25 (coefficient of y)
- F = -6 (constant term)
Let's calculate the discriminant using the formula:
D = B^2 - 4AC
Substituting the values from our equation:
D = (-4)^2 - 4(1)(0) = 16 - 0 = 16
Since D = 16 (positive and non-zero), the conic given by equation (b) is also an ellipse.
In conclusion, both equations (a) and (b) represent ellipses based on their discriminant values.