1) Use the general analytic equation to write specific equations for the ellipse and hyperbola. Sketch the resulting curves to confirm your answers.
Solns.
The general analytic equation is:
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
For ellipse: Let A = 9, B= 0, C = 4, D= 36, E = -8 and F = 4
Then we get: 9 x 2 + 4 y 2 + 36 x - 8 y + 4 = 0
With appropriate algebraic manipulation we get:
(x + 2) 2 / 4 + (y - 1 ) 2 / 9 = 1
Graphing yields:
For hyperbola: Let A = -1, B= 0, C = 4, D = E = F = 0
Then we get:
4 y 2 - x 2 + 4 x = 0
Graphing yields:
2) The discriminant: B2 - 4AC = B'2 - 4A'C'
has been found valid for any rotation of axes (i.e. for any angle Θ ) Thus, the quantity is invariant under a rotation of axes.
a) Show that for the parabola the discriminant: B2 - 4AC = 0
Solution: Take the sample parabola: y 2 = 4 x
The general analytic equation is:
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
Then for this case: B= 0 (no xy term), and A= 0, C = 1
Therefore: B2 - 4AC = - 4 AC = - 4(0)(1) = 0
b) Show that for the ellipse: B2 - 4AC < 0
Solution: Use the example ellipse of Problem (1):
9 x 2 + 4 y 2 + 36 x - 8 y + 4 = 0
B= 0, A = 9, C = 4
Therefore: B2 - 4AC = - 4 AC = - 4(9)(4) = -324
So the condition is satisfied
c) Show that for the hyperbola: B2 - 4AC > 0
Solution: Using the hyperbola from Problem 1:
4 y2 - x 2 + 4 x = 0
Therefore: B= 0, A = -1, C = 4
Then: B2 - 4AC = - 4 (-1)(4) = 16
Which is greater than zero so the condition holds.
3) Determine the equation to which:
x2 + xy + y2 = 1
reduces when the axes are rotated to eliminate the cross product term. Sketch the resulting curve.
And thereby identify the 2nd degree curve and graph it.
Solution:
The given equation has: A = B = C = 1
Choose Θ according to: cot 2Θ = (A - C)/ B = (1 - 1)/ 1 = 0/1 = 0
Then: 2Θ = 90 deg so: Θ = 45 deg
Then: x = (x' - y') / Ö 2
y = (x' + y' ) / Ö 2
And, after some algebra:
3(x') 2 + ( y') 2 = 2
Divide through by 2:
3(x') 2 / 2 + ( y') 2 / 2 = 1
This curve defines an ellipse which is shown below:
4) Use the discriminant to classify each of the following curves as a circle, parabola, ellipse, or hyperbola:
a) x2 + y2 + xy + x - y = 3
A = 1, B = 1, C = 1
Then: B2 - 4AC = (1)2 - 4 (1)(1) = 1 - 4 = -3
Since B2 - 4AC < 0 the curve is an ellipse
b) x2 - y2 = 1
A = 1, B = 0, C = -1
Then: B2 - 4AC = - 4 (1)(-1) = 4
Since B2 - 4AC > 0 the curve is a hyperbola
c) x2 + 3 y2 - 3 xy - 6 y = 7
A = 1, B = -3, C = 3
Then: B2 - 4AC = (-3)2 - 4(1) (3) = 9 - 12 = -3
Since B2 - 4AC < 0 the curve is an ellipse
d) x2 + y2 + 3 x - 2 y = 3
A = 1, B = 0, C = 1
Then: B2 - 4AC = - 4 (1)(1) = - 4
Since B2 - 4AC < 0 the curve is an ellipse
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