Find the vertex, focus, and directrix of parabola x^2+4x+2y=0

Find the vertex, focus, and directrix of parabola x^2+4x+2y=0


1 Answers

25757 Points
5 years ago
The initial step is to define what these terms mean. 

For a general quadratic we have the equation which defines a parabola. You should draw this on a graph to help you. 

a*x^2+b*x+c = 0 

vertex aka the point of greatest curvature. This may be found by taking the first derivatie and setting it to 0. Meaning it is where the rate of change of the curve hits 0 because the change is already at its highest.. That will be 2*a*x + b = 0 => x = -b/(2*a). 

focus aka a single point on the axis of symmetry. This and the directrix may be used to define a parabola aka generate the quadratic equation you got. The parabola i.e. the graph of your equation is such that at any point the sum of shortest distance of the point from focus + directerix is the same. 

directricx is a line perpendicular to the axis of symmetry and may be represented by x = p. 

For your equation re write it as the parabola below 

2*y = x^2 - 4*x 
y = x^2/2 - 2*x 
a = 1/2 
b = -2 
c = 0 

The vertex is at x = b/2a and y as defined by the equation so for vertex 
x = -b/2a = 2/1 = 2; y = 1/2 * 2^2 - 2*2 = 2-4 = -2

The focus goes throught the axis of symmetry and the vertex so it must have x =-b/2a = 2; we do not yet know the y. 

The directrix is a line so we just need to figure out its y. 

We denote the y for the foucs as "Y". 
The distance from the parabola to focus on the y axis is Y - 2. 
The distance from parabola to focus on the x axis is -b/2a or 2/2 = 1 
Plug in 
Y - 2 = 1 => Y = 3 
hence the focus is at x=2; y = 3 

Now for directrix we need to find its y on the axis of symmetry. There the distance of directrix from parabola is the same as that of the focus. 
y(parabola) - y(directrix) = y(focus) - y(parabola) => 
-2 - y(directrix) = 3 - 1 => 
y(directrix) = 2 - 2 
directrix is x = 0

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