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Author Question: Find an equation of a parabola satisfying the given conditions: Focus (2, 0) and directrix y = 2. (Read 65 times)

audragclark

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on: May 26, 2021

Question 1

Find an equation of a parabola satisfying the given conditions Focus (4, 1) and directrix x = -2.
◦ (y-1)2 = 12(x - 1)
◦ (y-1)2 =  -12(x - 1)
◦ y2 = 12x + 1
◦ (y+1)2 = 12(x + 1)
◦ y2 = 12x

Question 2

Find an equation of a parabola satisfying the given conditions: Focus (2, 0) and directrix y = 2π.
◦ x2 = 4π(1 - y)
◦ (x + 1)2 = 4π(π - y)
◦ (x - 2)2 = 4π(π - y)
◦ (x - 2)2 = 4(π - y)
◦ (x - 1)2 = 4π(π - y)
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Marked as best answer by audragclark on May 26, 2021

xthemafja

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Reply #1 on: May 26, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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audragclark

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Reply 2 on: May 26, 2021
:D TYSM

flexer1n1

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

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