Question

The equation $y^2+4x+4y+k=0$ represents a parabola whose latus rectum is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D

$4$

Explanation for the correct option:

Step 1: Standardize the given equation.

The equation $y^2+4x+4y+k=0$ of the parabola is given.

Standardize the given equation as follows:

Add and subtract $4$ from the Left-hand side.

$$\begin{gathered} y^2+4x+4y+k+4-4=0 \\ \Rightarrow \left(y^2+4y+4\right)+4x+k-4=0 \\ \Rightarrow {\left(y+2\right)}^2=4-4x-k \\ \Rightarrow {\left(y+2\right)}^2=-4\left[x+\frac{-4+k}{4}\right]...1 \end{gathered}$$

Therefore, the standardized equation of the parabola is ${(y+2)}^2=-4\left[x+\frac{-4+k}{5}\right]$.

Step 2: Find the length of the latus rectum.

We know that the general equation of a parabola is ${(y-y_1)}^2=-4a(x-x_1)...2$.

Where, $(x,y)$ is the general point of the parabola.

$(x_1,y_1)$ is the vertex of the parabola.

$4a$ is the length of the latus rectum.

On comparing equation $1$ and equation $2$, we get

The length of the latus rectum is $4$.

Hence, option $D$ is the correct .