Question

An equation of a parabola is

given.

(a) Find the focus, directrix, and focal diameter of the

parabola.

(b) Sketch a graph of the parabola and its directrix.

Short answer

a. The parabola $x+\frac{1}{5}{y}^2=0$has its focus at $\left(\frac{-5}{4},0\right)$. Its focal diameter is 5 and directrix is $x=\frac{5}{4}$.

b. The graph of parabola $x+\frac{1}{5}{y}^2=0$is

Step-by-step solution

a.Step 1. Given information.

The given equation of a parabola $x+\frac{1}{5}{y}^2=0$.

Step 2. Write the concept.

The given equation of a parabola $x+\frac{1}{5}{y}^2=0$

Putting the equation in standard form $y^2=4ax$

Thus we get, $x+\frac{1}{5}{y}^2=0$

$y^2=4\left(-\frac{5}{4}\right)x$

From the above equation $4a=-5$, so the focal diameter is 5

By solving $a=-\frac{5}{4}$ for a,

We get $a=-\frac{5}{4}$

$\therefore$ the focus is $\left(\frac{-5}{4},0\right)$ and the directrix is $x=\frac{5}{4}$

b.Step 1. Given information.

The given equation of a parabola $x+\frac{1}{5}{y}^2=0$.

Step 2. Write the concept.

The given equation of a parabola $x+\frac{1}{5}{y}^2=0$.

Converting the equation in standard form

Thus, $x+\frac{1}{5}{y}^2=0$ which represents a downward parabola with vertex at (0,0)

$y^2=-5x$

Hence the coordinates of the focus will be $\left(\frac{-5}{4},0\right)$

And the equation of the directrix will be $x=\frac{5}{4}$