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# Find the standard form of the equation of the parabola with a focus at (5, 0) and a directrix at x = -5.

I dont understand how to solve it. please explain the steps

given: x+5=0, focus (5,0)

ans→ we know that, eccentricity of a parabola is 1, and if PM is the line of directrix, point P is a moving point, and point S is the focus.
therefore, SP/PM=e i.e. SP=ePM.
SP^2= (h-5)^2+k^2 .....(using distance formula)
PM^2=(h+5)^2

Now, SP=ePM
(h-5)^2+k^2=(h+5)^2
By solving this, we get y^2=20x

### 4 Answers by Expert Tutors

Howard L. | Experienced Math Tutor Specializing in Test Prep & Grade Up SkillsExperienced Math Tutor Specializing in T...
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Let me add some more details abut the parabola before applying a formula,

1. Recognize the type, vertical(regular) or horizontal(sideways).
The difference in the two types is in which variable is squared:
for vertical parabolas, the x part is squared  and for horizontal, the y part is squared.
Here, for the question, the directrix is x=-5, it means that the graph type is horizontal.

2. Remember or you can derive the formula from y= ax2 + bx +c for the vertical, or x= ay2 + by + c for the horizontal,
such that, the vertex form of a parabola equation with its vertex at (h, k) as,
y = a(x – h)2 + k  → 1/a( y - k) = (x - h)2,   for the vertical
x = a(y – k)2 + h, →  1/a(x - h )= (y - k)2,   for the horizontal

Or, the conics form of the parabola equation as

4p(y – k) = (x – h)2, for the vertical
4p(x – h) = (y – k)2, for the horizontal

where 4p = 1/a, or a= 1/(4p).
Keep h with x, y with k, and p with the first degree part (not squared part)

The conic form is more useful because the variable p represents the distance between the vertex and the focus.  Also, the same p represents the distance from the vertex to the directrix.  The vertex is exactly midway between the directrix and the focus.  Obviously, 2p represents the distance between the focus and the directrix.

3. Find the variables for the formula recognized.
For the given question, the formula to be
4p(x – h) = (y – k)2, for the horizontal

From the focus (5,0) and the directrix is x=-5 given,
the value of h (which stays with x) is located in the middle of 5 and -5, that is 0,
and k (which stays with y) is the same y value of the focus, that is 0, therefore the vertex is at (0,0)
P is the distance of x-axix between the vertex and the focus, that is p = 5-0 = 5
Replace all the variables into the formula:  4(5)(x - 0) = (y - 0)2  →  20x = y2,  or y2 = 20x

I hope that this explanation helps to understand the parabola, and to remember its formula.  This is one of the formula easy to get confused and forget.

Stephen K. | Physics PhD experienced in teaching undergraduatesPhysics PhD experienced in teaching unde...
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Kate,

First of all if we sketch what we're given on a graph, we see that the directrix is a vertical line @ x = -5, and the focus @ (5,0) lies on the x-axis.  This tells us that our parabola is going to open to the right (+ x direction).  Next, the vertex has to lie half-way between the focus and the directrix, which means it will have coordinates x = 0, y =0 so the vertex is the point (0,0).

For a parabola opening to the right, the standard equation is given by:

(x-h)² = +4p(y-k)

h = x coordinate of the vertex = 0 in our case, k = y coordinate of the vertex also = 0.

p is the distance between the vertex and the focus = 5

So our equation is then (x-0)² = 4(5)(y-0)

Simplifying x² = 20y
SURENDRA K. | An experienced,patient & hardworking tutorAn experienced,patient & hardworking tut...
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Equation for parabola is

(y-k)^2=4p(x-h)

h= 0   ,  k =0

p =5

y^2=20x
Phillip R. | Top Notch Math and Science Tutoring from Brown Univ GradTop Notch Math and Science Tutoring from...
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You need to memorize the standard form or find it in a textbook.

For a parabola with a horizontal axis such as this example, the equation is

(y - k)2 = 4p(x - h)

(h , k) is the vertex

directrix equation is x = h - p

Hint: the vertex lies half way between the focus point and the directrix