Question

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

Answer 1

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.

$y = 12x^2+6x+24 \\ = 12(x^{2} + \frac{1}{2} x)+24 \\ = 12(x^{2} + \frac{1}{2} x+ \frac{1}{16}) +24- \frac{3}{4} \\ =12(x+ \frac{1}{4} )^2+ \frac{93}{4} \\ y-\frac{93}{4}=4(3)(x+ \frac{1}{4} )^2$

From the equation, the vertex is $(- \frac{1}{4} , \, \frac{93}{4} )$ 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 $y= \frac{93}{4}-3= \frac{81}{4}$

Therefore, the equation of the directrix is $y=\frac{81}{4}$


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 $x= -7$

Answer 2

Answer:  The equation of directrices are

(1) $y=\dfrac{81}{4}$

(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

$y=12x^2+6x+24\\\\\Rightarrow y=12\left(x^2+\dfrac{1}{2}x\right)+24\\\\\\\Rightarrow y=12\left(x^2+2\times x\times \dfrac{1}{4}+\dfrac{1}{16}\right)-\dfrac{12}{16}+24\\\\\\\Rightarrow y=12\left(x+\dfrac{1}{4}\right)^2-\dfrac{3}{4}+24\\\\\\\Rightarrow y=4\times 3\left(x+\dfrac{1}{4}\right)^2+\dfrac{93}{4}\\\\\\\Rightarrow y-\dfrac{93}{4}=4\times 3\left(x+\dfrac{1}{4}\right)^2.$

Comparing with the standard equation, we get

$k=\dfrac{93}{4},~~p=3.$

So, the equation of the directrix is

$y=k-p\\\\\Rightarrow y=\dfrac{93}{4}-3\\\\\\\Rightarrow y=\dfrac{81}{4}.$

(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

$(y-1)^2=16(x+3)\\\\\Rightarrow (y-1)^2=4\times4(x+3).$

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) $y=\dfrac{81}{4}$

(2) $x=-7.$