Question

: Write the equation of the dircetrix of the conic section shown below. Write your answer without using spaces.

Y^2 + 16y + 4x + 4 = 0

Answer:_________

Answer 1

Answer on Question #57357 – Math – Analytic Geometry

Question

Write the equation of the directrix of the conic section shown below. Write your answer without using spaces.

$$y^{2} + 16y + 4x + 4 = 0.$$

Solution

By definition the canonical equation of a parabola with horizontal axis of symmetry has the form

$$y^{2} = 2px,$$

where $p$ is the distance from the vertex to the focus and the vertex to the directrix. The equation of the directrix for parabola (1) has the form

$$x = -\frac{p}{2}.$$

The vertex of parabola (1) is at the point $(0, 0)$.

The conic section (*) is a parabola. Let's show it. Rewriting (1) in more convenient form we obtain

$$\begin{aligned} y^{2} + 16y + 64 - 64 + 4x + 4 &= 0 \\ (y + 8)^{2} + 4x - 60 &= 0 \\ (y + 8)^{2} &= -4 \cdot (x - 15). \end{aligned}$$

As we see, the equation (3) describes the parabola with the vertex at the point $(15, -8)$.

Let's introduce the new coordinates

$$y' = y + 8, \quad x' = x - 15.$$

Then the equation (3) takes the form

$$(y')^{2} = -4x'.$$

Comparing (2) and (5) we can write

$$2p = -4 \Rightarrow p = -2.$$

Thus, the equation of the directrix in the new system of coordinates (4) is

$$x' = -\frac{-2}{2} = 1$$ $$x' = 1.$$

Therefore, taking into account (4) and (6), we can write the directrix equation in the basic system of coordinates:

$$1 = x - 15$$ $$x = 16.$$

Answer: $x = 16$.

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