## claudineVeluya 2 years ago Hi! I'm Claudine Veluya a 2nd year college student. could someone have an additional information about parabola? thank you.

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1. UnkleRhaukus

2. claudineVeluya

I'm i right that Parabola is a set of all points equidistant from a line and a fixed point not on the line, the line is called the directrix, and the point is called the focus. The point on the parabola halfway between the focus and the directrix is the vertex. The line containing the focus and the vertex is the axis. A parabola is symmetric with respect to its axis.

3. sirm3d

also, the segment joining two points of the parabola, parallel to the directrix and through the focus is the latus rectum of the parabola.

4. claudineVeluya

the Latus rectum of the parabola is use if they have the given of vertex. I'm i right. sirm3d.

5. harsimran_hs4

well what i suggest is that you should refer some text for parabola as it will give you better understanding

6. UnkleRhaukus

7. claudineVeluya

oh its that an example of Parabola? unkleRhaukus

8. agent0smith

^ yep, any object thrown up into the air (other than straight up) follows a parabolic path due to gravity.

9. UnkleRhaukus

you can make a parabola by holding a string between two points

10. agent0smith

@UnkleRhaukus won't that make a catenary curve?

11. UnkleRhaukus

|dw:1361698911624:dw|

12. UnkleRhaukus

ok @agent0smith, your right , the hanging string is a catenary curve which is not quite the same as a parabola, but they look kinda similar

13. agent0smith

^ yep and you can model them with a parabola pretty closely.

14. claudineVeluya

i"m right that is an example of Property of Parabolas because The tangent line at a point P on a parabola makes equal with the line through P parallel to axis of symmetry and the line through P and the focus. let P |dw:1361757582217:dw|

15. claudineVeluya

hi! Do you think that it is way to derive the Parabola? DERIVATION OF PARABOLA A parabola coinciding was x-axis Its vertex at the origin Its focus at the (p,o) And its directix is x=-p Where p is called distance Then we choose any point P(x,y) By the definition of parabola FP=SP (√((x-p)^2+ y^2 ))^2= (x+p)^2 (x+p)^2+y^2= (x+p)^2 x^2- 2px+y^2= x^2+ 2px+ p^2 y^2= x^2+ 2px+p^2- x^2+ 2px- p^2 〖 y〗^2=4px DERIVATION OF PARABOLA A parabola coinciding was x-axis Its vertex at the origin Its focus at the (p,o) And its directix is x=-p Where p is called distance Then we choose any point P(x,y) By the definition of parabola FP=SP (√((x-p)^2+ y^2 ))^2= (x+p)^2 (x+p)^2+y^2= (x+p)^2 x^2- 2px+y^2= x^2+ 2px+ p^2 y^2= x^2+ 2px+p^2- x^2+ 2px- p^2 〖 y〗^2=4px can you give me other way of derivation of Parabola?