Question

Problem: A curve C is the locus of points in the xy-plane which are equidistant from the point (0,7.25 ) and the line (y=-7.25). Identify the curve?

Answer 1

This should scream PARABOLA!!!! It is the fundamental definition.

(0,7.25) is the Focus
y = - 7.25 is the Directrix

The vertex just might be at the Origin!

Answer 2

may be it has some sense but i cant understand why.

Answer 3

So, it didn't scream? Oh, well. Maybe we can just do the algebra.

equidistant
from the point (0,7.25 )
the line (y=-7.25)

Any old point might be described as (a,b).

How far is (a,b) from y = -7.25 = -29/4?

Answer 4

well, since point have coordinate (0, 7.25) then distance is \[D=\sqrt{a^2+(b-7.25)^2}\]

Answer 5

Good, you answered my next question. That is the distance from the point. We certainly need that. How about the other one. How far is (a,b) from y = -29/4?

Answer 6

well no clue how to find distance from (a,b) to line, since it can have different values.

Answer 7

b - (-29/4) -- On need consider only the y-value

Answer 8

how it give us a parabola then?

Answer 9

If you believe that last piece, we are set.

EQUIDISTANT

\(b + 7.25 = \sqrt{a^{2} + (b-7.25)^{2}}\)

Now, there's a little algebra to go. First, and foremost, we're talking about distances, so everything is positive and there is no harm in squaring things.

\((b + 7.25)^{2} = a^{2} + (b-7.25)^{2}\)

Is it looking better, yet?

Answer 10

now l start to geting something about it.

Answer 11

Okay, expand and simplify and you should see it!

Answer 12

thx you.

Answer 13

BTW - Excellent work coming up with that distance formula. This suggests you are paying attention. Medal for you!