Question

vv^(1) 3 4 5 6 7 Find an equation of the parabola with vertex (-3,1) and directrix y=6

          vv^(1) 3 4 5 6 7 Find an equation of the parabola with vertex (-3,1) and directrix y=6
        

Added by Daniel M.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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vv^(1) 3 4 5 6 7 Find an equation of the parabola with vertex (-3,1) and directrix y=6
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If a parabola's focus is at (3,4) and directrix is at y = 2, what is the vertex form of the equation representing this parabola? y = -1/12(x + 3)^2 - 1 y = 1/4(x - 3)^2 + 3 y = -1/12(x - 3)^2 - 1 y = 1/4(x + 3)^2 + 3

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Transcript

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00:01 Given the equation of parabola y is equal to minus of x plus one whole square plus six we compare this with general equation of parabola that is y equal to a x minus h whole square plus k we get a is equal to minus one h equal to minus one and k equal to six we know that the coordinate of the vertex is h comma k it means the vertex will be minus one comma six now we will find the value of p by comparing a with 1 divided by 4p.
00:37 We get minus 1 is equal to 1 divided by 4p.
00:42 It implies that p is equal to minus 1 divided by 4.
00:45 Now the coordinate of the vertex is h comma k plus p.
00:52 It is equal to minus 1 .6 minus 1 divided by 4...
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