Question

Derive the equation of the parabola with a focus at (6,2) and a directrix of y=1

Answer 1

Distance from focus to vertex=distance from vertex to directix

from (6,2) to y=1, the distance is 1
the focus is 1/2 higher than the vertex
vertex=(6,1.5)

for
(x-h)^2=4p(y-k)
vertex=(h,k)
p=distance from vertex to directix

vertex=(6,1.5)
ditance=0.5

(x-6)^2=4(0.5)(y-1.5)
(x-6)^2=2(y-1.5) is the equation
or
y=0.5x^2-6x+19.5 or
y=0.5(x-6)^2+1.5

Answer 2

Answer:

$(y-\frac{3}{2})^2=2(x-6)^2$  equation of parabola

Step-by-step explanation:

Focus is (h,k+p)=(6,2)

And directrix y=k-p=1

When comparing the values we get

h=6 and  

k+p=2    (a)

k-p=1      (b)

For  solving (a) and (b) we  substitute k=2-p in (b) we get:

$2-p-p=1$

$2-2p=1$

$p=\frac{1}{2}$

Hence, put $p=\frac{1}{2}$ in k=2-p we get:

$k=2-\frac{1}{2}$

$k=\frac{3}{2}$

Now, we have general equation as:

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

On substituting the values in general equation we get:

$(y-\frac{3}{2})^2=4\frac{1}{2}(x-6)^2$

$(y-\frac{3}{2})^2=2(x-6)^2$