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anonymous

  • 5 years ago

anyone help me another problem 2. Reduce each equation to standard form then find the coordinates of the center the foci, the end of the major axes and the ends of each latus rectum. Sketch the curve. Given problem: 4xsquared + ysquared +8x -4y-8=0

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  1. dumbcow
    • 5 years ago
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    Standard form: \[\frac{(x+1)^{2}}{4} + \frac{(y-2)^{2}}{16} = 1\] Center: (-1,2) Foci: (-1,2+-2sqrt(3)) Major axis: (-1,2 +-4) Minor axis: (-1 +-2, 2)

  2. anonymous
    • 5 years ago
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    thankyou dumbcow for helping me but i want to know how you get these answer. im scared if our prof: ask me where i get my answer then i cant answer him

  3. dumbcow
    • 5 years ago
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    First group x's and y's together and move constants on right hand side (4x^2 +8x) + (y^2-4y) = 8 Then use completing the square Factor the 4 out of x's part first 4(x^2+2x) +(y^2-4y) = 8 4((x+1)^2-1) +(y-2)^2-4 = 8 distribute 4 back in 4(x+1)^2 -4 +(y-2)^2 -4 = 8 move constants to right side 4(x+1)^2 + (y-2)^2 = 16 Now divide by 16 to get 1 on right side (x+1)^2 /4 + (y-2)^2 /16 = 1 Now you have it in standard form The rest is pretty straightforward once you have it in this form Use the attached reference sheet for ellipses to help you hope this helps

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