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pav-90 [236]
2 years ago
10

Which equation represents a parabola opening to the right with a vertex at the origin and focus at (9,0) A. X = - 1/36y^2 B. X =

- 1/6y^2 C. X = 1/36y^2 D. X = 1/6y^2
Mathematics
1 answer:
Virty [35]2 years ago
8 0

Answer:

Option C: x = (1/36)*y^2

Step-by-step explanation:

A horizontal parabola is written as

x = f(y) =  a*y^2 + b*y + c

If a is positive, the parabola opens to the right.

if a is negative, the parabola opens to the left.

And the vertex of the parabola has the y-value:

y = -b/2a

and the x value

x = f(-b/2a)

For our parabola, we know that:

Opens to the right:

Then the only options left are:

C) x = (1/36)*y^2

D) x = (1/6)*y^2

Because in both cases b = 0, both of the equations have the vertex in the point (0, 0).

Now let's see wich one has a focus at (9, 0)

If the vertex of our equation is:

(h, k)

Then the focus will be:

(h + a, k)

Where a is the directrix of the equation.

Here we know that the vertex is (0, 0) and the focus is (9, 0)

Then:

(0 + a, 0) = (9, 0)

The directrix is 9.

And the directrix is such that:

(x - h)^2 = 4*a*(y - k)

Replacing the values of h and k (both are zero) we get:

x^2 = 4*a*y

And we know that: a = 9

x^2 = 4*9*y

x^2 = 36*y

if we isolate y, we get:

(1/36)*x^2 = y

This is option C.

Then the correct option is C.  

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