MedVision ad

Locus question (1 Viewer)

ItDunMatter

New Member
Joined
Mar 2, 2009
Messages
4
Gender
Undisclosed
HSC
N/A
hello..

any help would be appreciated

Find the equation of the locus of a point which moves so that the sum of its distances from (0,-1) and (0,9) is always 5
 

anyscope

Member
Joined
May 18, 2009
Messages
70
Location
Sydney
Gender
Male
HSC
2010
eugh, I hate these questions. I can't help you, but you should try posting it in the Mathematics part of the forum. Lol.
 

study-freak

Bored of
Joined
Feb 8, 2008
Messages
1,133
Gender
Male
HSC
2009
hello..

any help would be appreciated

Find the equation of the locus of a point which moves so that the sum of its distances from (0,-1) and (0,9) is always 5
Logically thinking, the shortest distance from (0,-1) to (0,9) is 10 units. Hence sum of distances from those points to a point cannot be smaller than 10.

If the sum of distances>10, the paths of the point will describe an ellipse with major axis on the y-axis.
 

xV1P3R

Member
Joined
Jan 1, 2007
Messages
199
Gender
Male
HSC
2010
Let any point on the locus be P (x,y) so that your locus is all of these points.
Let your points (0,-1) be A and (0,9) be B

So the condition for your locus is PA + PB = 5

PA = rt[ (x - 0)² + (y- (-1))²] (distance formula)

= rt[x² + (y+1)²]

PB = rt[ (x - 0)² + (y - 9)²]

Sub these into PA + PB = 5

rt[x² + (y + 1)²] + rt[ (x - 0)² + (y - 9)²] = 5

You want to get rid of the roots, so you have to get rid of the square roots

rt[x² + (y + 1)²] = 5 - rt[ (x - 0)² + (y - 9)²]

x² + (y + 1)² = 25 - 2rt[ x² + (y - 9)²] + x² + (y - 9)²

0 + y² + 2y + 1 = 25 - 2rt[ x² + (y - 9)²] + y² - 18y + 81

20y + 1 - 25 - 81 = -2rt[ x² + (y - 9)²]

20y - 105 = -2rt[ x² + (y - 9)²]

(20y - 105)² = 4( x² + (y - 9)²)

400y² - 4200y + 11025 = 4x² + 4y² - 72y + 324

396y² - 4128y + 10701 = 4x²

Hmm I think I got something wrong above, cause this answer looks disgusting

Hope this helps
 

addikaye03

The A-Team
Joined
Nov 16, 2006
Messages
1,267
Location
Albury-Wodonga, NSW
Gender
Male
HSC
2008
Let any point on the locus be P (x,y) so that your locus is all of these points.
Let your points (0,-1) be A and (0,9) be B

So the condition for your locus is PA + PB = 5

PA = rt[ (x - 0)² + (y- (-1))²] (distance formula)

= rt[x² + (y+1)²]

PB = rt[ (x - 0)² + (y - 9)²]

Sub these into PA + PB = 5

rt[x² + (y + 1)²] + rt[ (x - 0)² + (y - 9)²] = 5

You want to get rid of the roots, so you have to get rid of the square roots

rt[x² + (y + 1)²] = 5 - rt[ (x - 0)² + (y - 9)²]

x² + (y + 1)² = 25 - 2rt[ x² + (y - 9)²] + x² + (y - 9)²

0 + y² + 2y + 1 = 25 - 2rt[ x² + (y - 9)²] + y² - 18y + 81

20y + 1 - 25 - 81 = -2rt[ x² + (y - 9)²]

20y - 105 = -2rt[ x² + (y - 9)²]

(20y - 105)² = 4( x² + (y - 9)²)

400y² - 4200y + 11025 = 4x² + 4y² - 72y + 324

396y² - 4128y + 10701 = 4x²

Hmm I think I got something wrong above, cause this answer looks disgusting

Hope this helps
nice try but i'm pretty sure that equation is symmetric about the line y=5, and its kinda just two huge parabolas

Type it into graphmatica and check it out yourself
 

iMAN2

Member
Joined
Mar 22, 2007
Messages
217
Gender
Male
HSC
N/A
Answer:

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

same as:
2 sqrt(x^2+y^2-18 y+81) = 5
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top