An equation that represents all points in the X-Y plane, equidistant from the point (1,1) & the line x = 5 ?

Question

These were all the possible solutions given in the back of the book............

a. y = x^2 / 4 - x/2 + 1/4

b. y = x^2 / 8 - x/4 - 7/8

c. x = -y^2 / 4 + y/2 + 7/4

d. y = -x^2 / 4 + x/2 + 7/4

e. y = -x^2 / 8 + x/4 + 23/8

f. x = y^2 / 4 - y/2 + 1/4

g. x = -y^2 / 8 + y/4 + 23/8

h. x = y^2 / 8 - y/4 - 7/8

i. none of these

Answer 1

You can get all this by plotting the point and line and thinking about points equidistant.
1) You know it is going to be a quadratic.
2) You know that it is going to be y² not x².
That leaves f, g, h, i

3) You know the point (3,1) is going to be on the line

f) x = 1²/4 - 1/2 + 1/4 = -1/2 ->wrong
g) x = -1²/8 + 1/4 +23/8 = 3 -> good choice
h) c = 1²/8 - 1/4 - 7/8 = -1 -> wrong

That leaves g and i, best bet is to look at the graphs to decide if g is the correct answer, i would say it is.

Answer 2

The locus of points equidistance from a point and a line is the definition of a parabola.

The directrix is a vertical line so the parabola opens sideways.

The directed distance from the directrix to the focus is 2p.

2p = 1 - 5 = -4
4p = -8

p is negative so the parabola opens to the left.

The vertex (h,k) is the midpoint between the focus and directrix.

(h,k) = [(1+5)/2, 1] = (3,1)

The equation for a horizontal parabola with center (h,k) is:

4p(x - h) = (y - k)²
-8(x - 3) = (y - 1)²
x - 3 = (-1/8)(y² - 2y + 1)
x = -y²/8 + y/4 + 23/8

The answer is g.

Answer 3

(x-1)^2+(y-1)2 =(x-5)^2
x^2-2x+1 +y^2-2y+1 = x^2-10x+25

8x= -y^2+2y+23
x=-y^2/8+y/4+23/8 (g)