Question

(x+2) ^2=-12 (y+1) find the vertex and directrix of parabola

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,
Easy question but a little lengthy,,,,,

FOR PARABOLA VERTEX::::::::

For a parabola vertex, given ,,

$(x + {2})^{2} = 12(y + 1)$

First off,,,, Isolating the given variable "y".

Switching the given sides.

$(x + {2})^{2} = 12(y + 1) \\ \\ 12(y + 1) = {(x + 2)}^{2}$

Dividing both the sides by 12 in the denominator,,,,

$\frac{12(y + 1)}{12} = \frac{ {(x + 2)}^{2} }{12}$

Simplifying the nominator and the denominator and subtracting "1" from both left and right hand sides,,,,

$y + 1 = \frac{ {(x + 2)}^{2} }{12} \\ \\ \\ y + 1 - 1 = \frac{ {(x + 2)}^{2} }{12} - 1 \\ \\ \\ y = \frac{ {(x + 2)}^{2} }{12} - 1$

Writing and hence rearranging it into the standard format of equation,, that is,,

$y = a {x}^{2} + bx + c$
$y = \frac{1 \times {x}^{2} }{12} + \frac{1 \times x}{3} + \frac{1}{3} - 1$

For a given parabola

$a {x}^{2} + bx + c$

The vertex's "x" equals to,,

$- \frac{ b}{2a}$

Here,,

$a = \frac{1}{12} \\ \\b = \frac{1}{3}$

Substituting the given values,,

$x = \frac{ ( - \frac{1}{3}) }{2 \times ( \frac{1}{12})}$

Applying the fraction rule of

$\frac{ - a}{b} = - ( \frac{a}{b} )$
$= - \frac{ \frac{1}{3} }{2 \times \frac{1}{12} }$

Applying fraction rule of

$\frac{ \frac{b}{c} }{a } = \frac{b}{c \times a}$

Therefore,,

$= \frac{ \frac{1}{3} }{2 \times \frac{1}{12} } = \frac{1}{3 \times 2 \times \frac{1}{12} } \\ \\ \\ = - \frac{1}{6 \times \frac{1}{12} } \\ \\ \\ = \frac{1}{ - \frac{1 \times 6}{12} } \\ \\ \\ = - \frac{1}{ \frac{6}{12} } \\ \\ \\ = - \frac{1}{ \frac{1}{2} } \\ \\ \\ = - \frac{2}{1} \\ \\ = - 2$

Therefore,,,,

$\boxed{x = - 2}$

Now putting the value of x = - 2 to find the value of "y".

$y = \frac{1 \times {( - 2)}^{2} }{12} + \frac{1 \times ( - 2)}{3} + \frac{1}{3} - 1 \\ \\ \\ y = \frac{ {2}^{2} }{12} + \frac{ - 1 \times 2 + 1}{3} - 1 \\ \\ \\ y = \frac{ {2}^{2} }{ {2}^{2} \times 3} - \frac{2 + 1}{3} - 1 \\ \\ \\ y = \frac{1}{3} - \frac{1}{3} - 1 \\ \\ \\ y = 0 - 1 \\ \\ \\ y = - 1$

Value of Y is,,,,

$\boxed{y = - 1}$

Therefore the parabola vertex is,,,,,

$\implies$

(- 2, -1)

If a < 0 , then the vertex is a maximum value .

If a > 0 , then the vertex is a minimum value.

Since ,,

$a = \frac{1}{12}$

Therefore ,,,,

Minimum (- 2 , -1)

PARABOLA DIRECTRIX::::::

Parabola directrix is where a parabola is the locus of points such that the distance to a point (the focus itself) equals the distance to a line ( the directrix).

The standard equation for up-down facing parabola with vertex "x" at (h , k) and a focal length | p | , is,,

$4p(y - k) = {(x - h)}^{2}$

Therefore,,,

$= {(x + 2)}^{2} = 12(y + 1) \\ \\ \\ = 12(y + 1) = {(x + 2)}^{2}$

By taking a factor of "12".

$= 12(y + \frac{12}{12} ) = {(x + 2)}^{2} \\ \\ \\ = 12(y + 1) = {(x + 2)}^{2}$

By a taking a factor of "4".

$4 \times \frac{12}{4} (y + 1) = {(x + 2)}^{2} \\ \\ \\ 4 \times 3(y + 1) = {(x + 2)}^{2} \\ \\ \\ 4 \times 3(y - ( - 1)) = {(x - ( - 2))}^{2}$

THERFORE,,,,,,,

(h , k) = (-2 , -1) , and , p = 3.

Parabola here is symmetric around the y - axis and so the directrix is a line parallel to the x - axis, a distance of "- p" from the center (-2 , -1) y - coordinate.

$y = - 1 - p \\ \\ \\ y = - 1 - 3 \\ \\ \\ y = - 4$

THERFORE,,,,

$\boxed{y = -4}$

HOPE IT HELPS AND SOLVES YOUR DOUBTS REGARDING THIS!!!!!