Question:
The equation of the directrix of the parabola $y^2 + 4y + 4x + 2 = 0$ is ______.
The equation of the directrix of the parabola $y^2 + 4y + 4x + 2 = 0$ is ______.
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For a shifted horizontal parabola $(y-k)^2 = 4a(x-h)$ (opening right), the directrix is $x = h - a$. For $(y-k)^2 = -4a(x-h)$ (opening left), the directrix is $x = h + a$. Always draw a quick sketch to confirm which direction the directrix is relative to the vertex!
Updated On: Jun 19, 2026
- $x = -1$
- $x = 1$
- $x = -\frac{3}{2}$
- $x = \frac{3}{2}$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We are given the general expanded equation of a shifted parabola. We must convert it into its standard vertex form to identify its orientation, vertex, and parameter '$a$', to ultimately find the equation of its directrix.
Step 2: Detailed Explanation:
The given equation is $y^2 + 4y + 4x + 2 = 0$.
To find the vertex form, we need to complete the square for the $y$ terms.
Group the $y$ terms together and move the $x$ terms to the right side:
$y^2 + 4y = -4x - 2$
Complete the square for $y^2 + 4y$:
Take half of the $y$-coefficient ($4/2 = 2$) and square it ($2^2 = 4$).
Add 4 to both sides of the equation:
$y^2 + 4y + 4 = -4x - 2 + 4$
$(y + 2)^2 = -4x + 2$
Factor out the coefficient of $x$ on the right side to get the standard form:
$(y + 2)^2 = -4(x - \frac{1}{2})$
This equation perfectly matches the standard shifted left-opening parabola format:
$(y - k)^2 = -4a(x - h)$
By direct comparison, we extract the structural parameters:
Vertex $(h, k) = (\frac{1}{2}, -2)$
$4a = 4 \implies a = 1$
Because the $y$ term is squared and the right side is negative, this parabola opens horizontally to the left.
For a standard left-opening parabola $Y^2 = -4aX$ centered at the origin, the directrix is the vertical line located "behind" the vertex at $X = +a$.
When the vertex is shifted horizontally to $h$, the directrix line shifts with it.
The equation for the directrix becomes:
$x = h + a$
Substitute the values we extracted:
$x = \frac{1}{2} + 1$
$x = \frac{3}{2}$
Step 3: Final Answer:
The equation of the directrix is $x = \frac{3}{2}$, matching option (d).
We are given the general expanded equation of a shifted parabola. We must convert it into its standard vertex form to identify its orientation, vertex, and parameter '$a$', to ultimately find the equation of its directrix.
Step 2: Detailed Explanation:
The given equation is $y^2 + 4y + 4x + 2 = 0$.
To find the vertex form, we need to complete the square for the $y$ terms.
Group the $y$ terms together and move the $x$ terms to the right side:
$y^2 + 4y = -4x - 2$
Complete the square for $y^2 + 4y$:
Take half of the $y$-coefficient ($4/2 = 2$) and square it ($2^2 = 4$).
Add 4 to both sides of the equation:
$y^2 + 4y + 4 = -4x - 2 + 4$
$(y + 2)^2 = -4x + 2$
Factor out the coefficient of $x$ on the right side to get the standard form:
$(y + 2)^2 = -4(x - \frac{1}{2})$
This equation perfectly matches the standard shifted left-opening parabola format:
$(y - k)^2 = -4a(x - h)$
By direct comparison, we extract the structural parameters:
Vertex $(h, k) = (\frac{1}{2}, -2)$
$4a = 4 \implies a = 1$
Because the $y$ term is squared and the right side is negative, this parabola opens horizontally to the left.
For a standard left-opening parabola $Y^2 = -4aX$ centered at the origin, the directrix is the vertical line located "behind" the vertex at $X = +a$.
When the vertex is shifted horizontally to $h$, the directrix line shifts with it.
The equation for the directrix becomes:
$x = h + a$
Substitute the values we extracted:
$x = \frac{1}{2} + 1$
$x = \frac{3}{2}$
Step 3: Final Answer:
The equation of the directrix is $x = \frac{3}{2}$, matching option (d).
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