Answer:
Given: In triangle ABC , AD is drawn perpendicular to BC.
Since AD is drawn perpendicular to BC, it creates two right triangles: ADB and ADC.
Prove that: 
Pythagoras triangle for right angle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.
In a right angle triangle ADB;
[By Pythagoras theorem]
or
.......[1]
Now, in right angle triangle ADC;
[By Pythagoras theorem]
or we can write this as;
......[2]
Substituting the equation [1] in [2] we get;
hence proved!
<span>y=a((x-h)^2)+k
Vertex=(h,k)
1. Vertex (5,-1) Point (2,4)
y=a((x-5)^2)+(-1)
f(2)=4
4=a((2-5)^2)+(-1)
4=a(-3)^2-1
4=a*9-1
5=a*9
5/9=a
y=(5/9)((x-5)^2)-1
2. Vertex (-2,0) Point (-1,-7)
y=a((x+2)^2)+0
y=a((x+2)^2)
f(-1)=-7
-7=a(-1+2)^2
-7=a(1)^2
a=-7
y=-7(x+2)
(x-h)^2=4(d)(y-k)
3. Vertex (0,0) Focus (0,2)
d=f-v
d=2
(x^2)=4(2)(y)
(x^2)=8y
f(0)=0
4. Focus (-3,4) Directrix y= -2
(x-h)^2=4(d)(y-k)
d=(4-(-2))/2=6/2=3
(x-h)^2=4(3)(y-k)
h=-3
k=(-2+4)/2=(2)/2=1
(x+3)^2=4(3)(y-1)
(x+3)^2=12(y-1)</span>
Are you sure the equation does say 2X+16=(-1)?
