Question

How do you find the vertex, focus and directrix of ${y^2} + 4y + 8x - 12 = 0$ ?

Answer 1

In order to this question, to find the vertex, focus and the directrix of the given equation, firstly we will separate both the variables in L.H.S and R.H.S and then we will simplify the given equation by complete squaring both sides to make one of the standard equation of any curve. And then we will apply the formula to find the vertex focus and directrix according to the curve types.

Complete step by step answer:
Given equation: ${y^2} + 4y + 8x - 12 = 0$
We can also write the given equation as: ${y^2} + 4y = - 8x + 12$
Now, we will completing the square by adding 4 in both sides of the equation:-
${y^2} + 4y + 4 = 4 - 8x + 12 \\\ \Rightarrow {(y + 2)^2} = - 8x + 16 = 4.( - 2)(x - 2)$
Now, we will make the above equation in standard equation-
Let, $(y + 2) = Y$ and $(x - 2) = X$ , so the equation becomes:
${Y^2} = 4aX$ ……….(i)
The above equation is a std. equation of a Parabola, with $a = - 2$.

Now, as we know the parabola, the Vertex is $(0,0)$, Focus is $S(a,0)$ and the equation of the directrix is $X + a = 0$ .
To find the Vertex:
$(X,Y) = (0,0) \\\ \Rightarrow (x - 2,y + 2) = (0,0) \\\ \therefore (x,y) = (2, - 2)$

To find the Focus:
$S(X,Y) = S(a,0) = S( - 2,0) \\\ \Rightarrow S(x - 2,y + 2) = S( - 2,0) \\\ \therefore S(x,y) = S(0, - 2)$

To find the Directrix:
Equation is:
$X + a = 0 \\\ \Rightarrow x - 2 = 0 \\\ \Rightarrow x - 2 - 2 = 0 \\\ \therefore x - 4 = 0 \\\$
Hence, the required vertex is $(2, - 2)$ , focus is $(0, - 2)$ and the equation of the directrix is $x - 4 = 0$.

Note: Every point on the parabola is the same distance from the directrix and the focus (equidistant). Line 1 from the directrix to the parabola is the same length as line 1 from the parabola back to the focus, in other words. The same is true for all other distances between a parabola point and the focus and directrix.