Question

The equation of a parabola is given. y=12x2+6x+24 What is the equation of the directrix of the parabola?

The equation of a parabola is (y−1)2=16(x+3) .
What is the equation of the directrix of the parabola?

Answer 1

Answer: The equation of directrices are

(1)
`y=(81)/(4)`

(2)
`x=-7.`

Step-by-step explanation: We are given to find the equation of the directrices of the following parabolas:

`y=12x^2+6x+24~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\(y-1)^2=16(x+3)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)`

(1) The equation of the directrix for the parabola
`(y-k)=4p(x-h)^2,` is given by

`y=k-p.`

From equation (i), we have

`y=12x^2+6x+24\\\\\Rightarrow y=12\left(x^2+(1)/(2)x\right)+24\\\\\\\Rightarrow y=12\left(x^2+2* x* (1)/(4)+(1)/(16)\right)-(12)/(16)+24\\\\\\\Rightarrow y=12\left(x+(1)/(4)\right)^2-(3)/(4)+24\\\\\\\Rightarrow y=4* 3\left(x+(1)/(4)\right)^2+(93)/(4)\\\\\\\Rightarrow y-(93)/(4)=4* 3\left(x+(1)/(4)\right)^2.`

Comparing with the standard equation, we get

`k=(93)/(4),~~p=3.`

So, the equation of the directrix is

`y=k-p\\\\\Rightarrow y=(93)/(4)-3\\\\\\\Rightarrow y=(81)/(4).`

(2) The equation of the directrix for the parabola
`(y-k)^2=4p(x-h),` is given by

`x=h-p.`

From equation (i), we have

`(y-1)^2=16(x+3)\\\\\Rightarrow (y-1)^2=4*4(x+3).`

Comparing with the standard equation, we get

`h=-3,~~p=4.`

So, the equation of the directrix is

`x=h-p\\\\\Rightarrow x=-3-4\\\\\Rightarrow x=-7.`

Thus, the equation of directrices are

(1)
`y=(81)/(4)`

(2)
`x=-7.`

Answer 2

1.) Given the equation of a parabola

`y = 12x^2+6x+24`

The vertex form of a parabola is given by

`y-k=a(x-h)^2=4p(x-h)^2`
where: (h, k) is the vertex and p is the distance between the vertex and the directrix.


`y = 12x^2+6x+24 \\ = 12(x^(2) + (1)/(2) x)+24 \\ = 12(x^(2) + (1)/(2) x+ (1)/(16)) +24- (3)/(4) \\ =12(x+ (1)/(4) )^2+ (93)/(4) \\ y-(93)/(4)=4(3)(x+ (1)/(4) )^2`

From the equation, the vertex is
`(- (1)/(4) , \, (93)/(4) )` and the distance between the vertex and the directrix is 3.

Because, the x-part of the equation is squared and the value of p is positive, this means that the parabola opens up and the directrix is a horizontal line having the value y = c, where c is the y-value of the vertex - 3

Equation of the directrix is
`y= (93)/(4)-3= (81)/(4)`

Therefore, the equation of the directrix is
`y=(81)/(4)`


2.) Given the equation of a parabola written in vertex form

`(y-1)^2 = 16(x+3)`

The vertex form of a parabola is given by

`(y-k)^2=a(x-h)=4p(x-h)`
where: (h, k) is the vertex and p is the distance between the vertex and the directrix.


`(y-1)^2 = 16(x+3)=4(4)(x+3)`

From the equation, the vertex is
`(-3 , \, 1)` and the distance between the vertex and the directrix is 4.

Because, the y-part of the equation is squared and the value of p is positive, this means that the parabola opens to the right and the directrix is a vertical line having the value x = c, where c is the x-value of the vertex - 4

Equation of the directrix is
`x= -3-4= -7`
Therefore, the equation of the directrix is
`x= -7`