HELP studying for exam.. LOCUS? (1 Viewer)

[mobilephone99]

I've attempted this question but not sure if I'm right..

Q) Derive the locus of P(x,y) such that its distance from A(-1,5) is equal to its perpendicular distance from the line y=9.

I tried doing this:

Let B lie on the line y=9.
PA = PB therefore PA^2 = PB^2

(x+1)^2 + (y-5)^2 = (x-x)^2 + (y-9)^2

x^2 + 2x + 1 + y^2 - 10y + 25 = x^2 - 2x^2 + x^2 + y^2 - 18y + 81

x^2 + 2x + 8y - 55 = 0 is the locus.

Any suggestions for a different solution to the question or have I figured it out?

I'd be very thankful for any help, as my exam is tomorrow :S

 

[gurmies]

√[(x+1)^2 + (y-5)^2] = |9-y|

(x+1)^2 + (y-5)^2 = (9-y)^2

x^2 + 2x + 1 + y^2 - 10y + 25 = 81 - 18y + y^2

x^2 + 2x + 1 + y^2 - 10y + 25 - 81 + 18y - y^2 = 0

x^2 + 2x - 55 + 8y = 0

(x+1)^2 - 1 - 55 + 8y = 0

(x+1)^2 = -8y + 56

(x+1)^2 = -8(y-7)

Therefore, Parabola is concave down, has a focal length of 2 and has a vertex of (-1, 7)

 

[mobilephone99]

Thank you so much!
But I'm just wondering why it equals |9-y|?
The rest I understand...
If you could explain that to me it'd be a great help lol.

 

[gurmies]

Well, the best explanation I can provide, is that's the distance of the point (x,y) from the line y = 9. Since we want the perpendicular distance, there will be no horizontal component. Looking back at our definition of the absolute value, it's defined as the distance from something, in this case the line y= 9. Your way is perfectly legitimate as well, I just like using absolutes =)

 

[mobilephone99]

Ah ok, so my answer was correct after all, haha. I realised that after I expanded your answer out and it was the same as mine.

Thanks a million

 

[gurmies]

No problem =D, glad you understand!