230k views
0 votes
What is the equation of the quadratic graph with a focus of (3,1) and a directrix of y=5

What is the equation of the quadratic graph with a focus of (3,1) and a directrix-example-1

2 Answers

4 votes

Answer:

b

Step-by-step explanation:

User Crosbie
by
8.3k points
7 votes

Answer:


y = -(1)/(8) (x-3)^2 + 3

Step-by-step explanation:

Assume a general point (x,y) that belong to the curve

To get the equation of the parabola, we follow the given steps:

1- Find the distance between point (x,y) and the focus point:

The general point is (x,y) and the focus is (3,1)

Using the distance formula:

D =
√((x-3)^2+(y-1)^2) .................> equation I

2- Find the distance between the point (x,y) and the directrix:

Directrix is a horizontal line having the general equation y=c

We are given that the directrix is y = 5

Therefore, the distance between the point and the directrix will simply be:

|y-c| = |y-5| ...............> equation II

3- Equate equations I and II and solve for y:


√((x-3)^2+(y-1)^2) = |y -5|

a- Square both sides:

(x-3)² + (y-1)² = (y-5)²

b- Now, expand the brackets having y:

(x-3)² + y² - 2y + 1 = y² - 10y + 25

c- Finally, solve for y:

(x-3)² + 1 - 25 = -10y + 2y

(x-3)² - 24 = -8y


y = -(1)/(8) (x-3)^2 + 3

Attached is the graph for the parabola.

Hope this helps :)

What is the equation of the quadratic graph with a focus of (3,1) and a directrix-example-1
User MonkeyWidget
by
8.0k points