What is the vertex form of the equation of the parabola with a focus at (15,-5) and a directrix of #y=7 #?

Question

Eva Adamson · Accepted Answer

Conic form: #(x-11)^2 = 16(y-5)# We know this up/down parabola #=># equation of directrix is #y=7# General equation #(x-h)^2 =4p(y-k)# Focus #(h+p, k)# Directrix: #y= h-p# Focus #(15, -5)# so 1) #h+p= 15# ; #color(red)(k= -5)# Directrix: #y= 7# 2) #h-p=7# Now we have system of equation (add both equations) # h+p= 15# #h--p = 7# #=> 2h =22 =># #color(red)(h= 11)# Now we solve for #p# #h+p= 15# #(11)+p =15# #=> color(red)(p= 4)# Conic form #(x-color(red)11)^2 =4*color(red)(4)(y-color(red)5)# Answer: #(x-11)^2 = 16(y-5)#

Nathan Axel · Answer

undefined The vertex form of the equation of a parabola with a focus at $(h, k + p)$ and a directrix of $y = k - p$ is given by:

$$ (x - h)^2 = 4p(y - k) $$

For the given focus at $(15, -5)$ and directrix $y = 7$, we have:

$$ h = 15, \ k = \frac{{7 - 5}}{2} = 6, \ p = 6 - (-5) = 11 $$

Substituting these values into the vertex form equation:

$$ (x - 15)^2 = 4 \cdot 11(y - 6) $$