Question 1) Vertex of the Parabola.![y-3=12(x+5)^(2) \\ \\ y= 12(x+5)^(2)+3](https://img.qammunity.org/2019/formulas/mathematics/high-school/zrm4oeduyl86ege1mcohlv9wxyih3rrd3a.png)
The vertex of the general parabolic equation:
![y=a(x-h)^(2)+k](https://img.qammunity.org/2019/formulas/mathematics/high-school/e2bunh2ysii185x4929mykdryamoip2aib.png)
lies at (h,k)
Comparing the given equation to general equation, we can write:
h = - 5
k = 3
So, the vertex of the given parabola will be (-5, 3)Question 2) Equation of DirectrixThe correct equation of the parabola is:
![y= (1)/(2) x^(2) +6x+23 \\ \\](https://img.qammunity.org/2019/formulas/mathematics/high-school/bui9dicun7egwkhwprm0cpinp5ic1x5q4e.png)
First we need to convert the equation to standard form as shown below:
![y= (1)/(2)( x^(2) +12x)+23 \\ \\ y= (1)/(2)( x^(2) +12x+36)+23- (1)/(2)(36) \\ \\ y= (1)/(2)(x+6)^(2)+5 \\ \\ y-5= (1)/(2)(x+6)^(2) \\ \\ 2(y-5)=(x+6)^(2) \\ \\ 4( (1)/(2))(y-5)= (x+6)^(2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/hyx0bk6ibog16d6s8shqv04w8p84ynz3jz.png)
The directrix of the general parabola of the form:
![4p(y-k)=(x-h)^(2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/gga5rk5dmegekpzczvuuysuywa738rqv8i.png)
is
y = k - pComparing equation of given parabola with general parabolic equation, we can write:
p =1/2
k = 5
So, equation of directirx will be:
y = 5 - 1/2 = 4.5So, option B gives the correct answer.Question 3) Focus of the parabolaThe correct equation of the parabola is:
![(1)/(32)(y-2)^(2)=x-1 \\ \\ (y-2)^(2)=32(x-1) \\ \\ (y-2)^(2)=4*8(x-1)](https://img.qammunity.org/2019/formulas/mathematics/high-school/r4kvpuie4hwxigre09wij5muljvwjj74b3.png)
Comparing this equation to the general parabolic equation, we can write:
p=8
h =1
k = 2
The focus of the parabola will be at
(h+p,k) = (9,2)So the focus of the parabola is at (9,2)
Thus, option D gives the correct answer.