Question

48. In the parabola, y = 3x ^ 2 + 12x + 11 focus is located at a distance p > 0 from the vertex. Then p=a. 3b. 1/3c. 12d. 1/12e. None of the above

Answer 1

Given the equation,

`y=3x^2+12x_{}+11`

We are to solve for the vertex first, in order to solve for the vertex.

`3x^2+12x+11=y`

factor all through by 3

`\begin{gathered} (3x^2)/(3)+(12x)/(3)+(11)/(3)=y \\ 3(x^2+4x+(11)/(3))=y\ldots\ldots.1 \end{gathered}`
`x^2+4x=-(11)/(3)\text{ complete the square for the inner expression}`
`\begin{gathered} x^2+4x+((4)/(2))^2=-(11)/(3)+((4)/(2))^2 \\ (x+2)^2=-(11)/(3)+4=(1)/(3) \\ =(x+2)^2-(1)/(3) \end{gathered}`

Put (x+2)²-1/3 into equation 1

`3((x+2)^2-(1)/(3))=y\ldots\ldots2`

The vertex is at (-2,-1)

Note:

`\begin{gathered} \text{vertex}=(h,k) \\ \text{focus}=(h,k+(1)/(4a)) \end{gathered}`

P is the distance between the focus and the vertex.

`\begin{gathered} (h-h,k+(1)/(4a)-k)=(0,(1)/(4a)) \\ \end{gathered}`

where,

`a=3\text{ from equation 2}`

Therefore,

`\begin{gathered} p=(0,(1)/(4*3))=(0,(1)/(12)) \\ p=(0,(1)/(12)) \end{gathered}`

Hence,

`p=(1)/(12)`

The correct answer is 1/12 [option D].