I'm being asked to graph a system of equations and I shouldn't have any trouble doing that if I can identify which type of conics the equations represent, but I'm having a bit of trouble doing that. Here's the equations:
(1) 5x^2 - 5y^2 = -35
(2) 5x^2 + 2y^2 = 77
I'm pretty sure eq. (1) represents the hyperbola y^2/7 - x^2/7 = 1, but I'm not positive. Still a bit clueless for eq. (2).
Input appreciated.
Also, if anyone knows how to graph conics on a ti-84 calculator that would help too, thanks.
(1) 5x^2 - 5y^2 = -35
(2) 5x^2 + 2y^2 = 77
I'm pretty sure eq. (1) represents the hyperbola y^2/7 - x^2/7 = 1, but I'm not positive. Still a bit clueless for eq. (2).
Input appreciated.
Also, if anyone knows how to graph conics on a ti-84 calculator that would help too, thanks.
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Equation #1)
5x^(2) - 5y^(2) = - 35
Divide both sides by - 35.
[ (- x^(2) ) / (7) ] + [ (y^(2) ) / (7) ] = 1
Yes this is a hyperbola with the transverse axis being vertical.
Center is at (h, k) ---------> (0, 0) where h = 0 and k = 0 at the origin
a^(2) = 7, b^(2) = 7
a = sqrt(7),
b = sqrt(7)
Foci: c^(2) = a^(2) + b^(2)
c^(2) = 7 + 7
c^(2) = 14
c = +/- sqrt (14)
Vertices are at: (h, k+a), (h, k - a)
Vertices -------> (0, sqrt (7) ), (0, - sqrt (7) )
Foci: (h, k + c), (h, k - c)
Foci -----------> (0, sqrt(14) ), (0, - sqrt(14) )
Eccentricity ------> e = c /a = [sqrt(14) ] / [7]
Ends of the conjugate axis are at:
(h +b, k), (h - b, k) ----------------------------> (sqrt(7), 0), ( - sqrt(7), 0)
Asymptotes of a Hyperbola
y = k + (a/b) * (x - h), ----------------> y = 0 + (1) * (x - 0) ---------> y = x
y = k - (a/b) * (x - h) -------------------> y = 0 - 1 * (x - 0) ------------> y = - x
Equation #2)
5x^(2) + 2y^(2) = 77
Divide throughout the equation by 77.
[(5x^(2)) / (77)] + [(2y^(2)) / (77)] = 1
Equation is an ellipse
Center of ellipse (h, k) where h = 0, k = 0 is at the origin ----> (0, 0)
b^(2) = 77 / 5 --------------------> b = +/- sqrt(77/5)
a^(2) = 77/2 ---------------------> a = +/- sqrt(77/2)
Foci: c^(2) = a^(2) - b^(2)
c^(2) = (77 / 2) ^(2) - (77 / 5) ^(2)
c^(2) = 5929/4 - 5929/25 = 124509/ 100
c = +/- sqrt(124509/100) = +/- 35.28583285
Major axis is vertical
5x^(2) - 5y^(2) = - 35
Divide both sides by - 35.
[ (- x^(2) ) / (7) ] + [ (y^(2) ) / (7) ] = 1
Yes this is a hyperbola with the transverse axis being vertical.
Center is at (h, k) ---------> (0, 0) where h = 0 and k = 0 at the origin
a^(2) = 7, b^(2) = 7
a = sqrt(7),
b = sqrt(7)
Foci: c^(2) = a^(2) + b^(2)
c^(2) = 7 + 7
c^(2) = 14
c = +/- sqrt (14)
Vertices are at: (h, k+a), (h, k - a)
Vertices -------> (0, sqrt (7) ), (0, - sqrt (7) )
Foci: (h, k + c), (h, k - c)
Foci -----------> (0, sqrt(14) ), (0, - sqrt(14) )
Eccentricity ------> e = c /a = [sqrt(14) ] / [7]
Ends of the conjugate axis are at:
(h +b, k), (h - b, k) ----------------------------> (sqrt(7), 0), ( - sqrt(7), 0)
Asymptotes of a Hyperbola
y = k + (a/b) * (x - h), ----------------> y = 0 + (1) * (x - 0) ---------> y = x
y = k - (a/b) * (x - h) -------------------> y = 0 - 1 * (x - 0) ------------> y = - x
Equation #2)
5x^(2) + 2y^(2) = 77
Divide throughout the equation by 77.
[(5x^(2)) / (77)] + [(2y^(2)) / (77)] = 1
Equation is an ellipse
Center of ellipse (h, k) where h = 0, k = 0 is at the origin ----> (0, 0)
b^(2) = 77 / 5 --------------------> b = +/- sqrt(77/5)
a^(2) = 77/2 ---------------------> a = +/- sqrt(77/2)
Foci: c^(2) = a^(2) - b^(2)
c^(2) = (77 / 2) ^(2) - (77 / 5) ^(2)
c^(2) = 5929/4 - 5929/25 = 124509/ 100
c = +/- sqrt(124509/100) = +/- 35.28583285
Major axis is vertical
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Given: 5x² − 5y² = -35
Add 5y² and 35: 5y² = 5x² + 35
Divide by 5: y² = x² + 7
Solve: y = ±√(x² + 7)
Graphing that on your TI-84 Plus will give you your hyperbola.
The second equation is an ellipse.
Add 5y² and 35: 5y² = 5x² + 35
Divide by 5: y² = x² + 7
Solve: y = ±√(x² + 7)
Graphing that on your TI-84 Plus will give you your hyperbola.
The second equation is an ellipse.