Question

The point P(x,y) moves so that the sum of the squares of tis distances from (4,1) and (2,-5) is 45. Find the equation of the locus

Answer 1

a sum of squares defines a circle; or perhaps an elipse

Answer 2

(4-2)^2 + (1--5)^2 = 45

4 + 36 = 45 ; /45

4/45 + 36/45 = 1 ; need a constant of variation somplace :)

Answer 3

the equation of the locus? i assume that means its center, or dies it mean foci?

Answer 4

(4,1) is one foci; (2,-5) is the other foci; this is an ellipse that is tilted

Answer 5

the tilt is the angle from foci to foci

Answer 6

slope = 6/2 = 3; which might be useful, tan(t) = 3

But the distance between the foci divided by 2 tells us the center point, right?

sqrt(2^2 + 6^2); sqrt(4+36); sqrt(40) = 2sqrt(10)

d/2 = sqrt(10) between the foci :)

Answer 7

wow...am so confused my lecturer told us to skip this question but...i just wanted to get a feel on how the working would look..:|

Answer 8

halfway between 2 and 4 is 3; so its centerd at x=3

halfway between -5 and 1 is -2; so its centered at y=-2

(x-3)^2 (y+2)^2
------ + ------- = 1 should be a general set up
a^2(45) b^2(45)

Answer 9

oh ok makes a little more sense now thanks so much amistre!

Answer 10

the titled/rotated conics are pretty cool, but I aint got all the info in my head to do them :)

Answer 11

thats the double intercept formula that you used?

Answer 12

okay...lol

Answer 13

i got no idea what the formula is called :) but yes, its the one that makes an ellipse ;)

Answer 14

okay cool thanks again so much really do appreciate amistre!

Answer 15

just in case:

\[\frac{(x-3)^2}{(60-20\sqrt{3})}+\frac{(y+2)^2}{(50-20\sqrt{3})}=1; \text{tilted at an angle of: a = }tan^{-1}(3)\]

Answer 16

\(\frac{(x-3)^2}{(60-20\sqrt{3})}+\frac{(y+2)^2}{(50-20\sqrt{3})}=1; \text{tilted at an angle of: a = }tan^{-1}(3)\)

Answer 17

lol .... titled at an angle of: a = tan-1(3)

Answer 18

okay