
Mark M. answered 12/02/20
Mathematics Teacher - NCLB Highly Qualified
y = a(x - h)2 + k
y = (1/16)(x - 2)2 - 1)
vertex at (h, k) or (2, -1)
Directrix:
y = k - p
p = 1 / 4a
p = 1 / (4)(1/16)
p = 4
y = -1 - 4
y = -5
Paul S.
asked 12/01/20Find the directrix for the parabola
Mark M. answered 12/02/20
Mathematics Teacher - NCLB Highly Qualified
y = a(x - h)2 + k
y = (1/16)(x - 2)2 - 1)
vertex at (h, k) or (2, -1)
Directrix:
y = k - p
p = 1 / 4a
p = 1 / (4)(1/16)
p = 4
y = -1 - 4
y = -5
Dayv O. answered 12/01/20
Attentive Reliable Knowledgeable Math Tutor
right off we see the vertex is (2,-1). at x=2 the distance from vertex to focus and vertex to directix does not involve x when at the vertex.
very generally when it is set that any point distance from focus and directix are equal,
the equation is ((x-xf)2)/(yf-yd)=y-(yf+yd), a parabola.
yf=y focus coodinate
yd= y directix coordinate
two equations simultaneous
yf+yd=-1
yf-yd=16
yd=-17/2 straight line horizopntal to x-axis
Paul S.
Ohh okay thank you tho12/02/20
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Dayv O.
yes, I checked my work and something did not add up. general equation, ((x-xfocus)^2/(2*yfocu-2*ydirectix))=y-(yfocus/2+ydirectix/2) simultaneous equations now (yf/2)+(yd/2)=-1 2*yf-2*yd=16 and yes ydirectix is y=-5 and yfocus =3 so at vertex, x=2, the distance from directix to vertex and distance from focus to vertex are equal which is the deirvation/definition of a parabola.12/02/20