conics question (1 Viewer)

[c0okies]

hi i'd like to ask for any help available for the last part of this question.. i've tried everything but i still cant show the answer

1. the normal at a point p(x1,y1) on the ellipse x^2/16 + y^2/9 = 1 cuts the x-axis and y-axis at two points A and B respectively

a) SHow the equation of the normal PB is 16x/x1 - 9y/y1 = 7

b) Then find the coordinates of point A and point B in terms of x1 and y1. Show that the coordinates of midpoint M of AB is (7x1/32, -7x1/18)

c) Show that the locus of the set of midpoints M of the interval AB is an ellipse with the same eccentricity as the original ellipse x^2/16 + y^2/9 =1 <==== cant do this one

 

[Mountain.Dew]

hi i'd like to ask for any help available for the last part of this question.. i've tried everything but i still cant show the answer

1. the normal at a point p(x1,y1) on the ellipse x^2/16 + y^2/9 = 1 cuts the x-axis and y-axis at two points A and B respectively

a) SHow the equation of the normal PB is 16x/x1 - 9y/y1 = 7

b) Then find the coordinates of point A and point B in terms of x1 and y1. Show that the coordinates of midpoint M of AB is (7x1/32, -7x1/18)

c) Show that the locus of the set of midpoints M of the interval AB is an ellipse with the same eccentricity as the original ellipse x^2/16 + y^2/9 =1 <==== cant do this one

have u tried to substitute the midpt M into the normal PB?
try subbing x=7x1/32 and y = -7x1/18 into 16x/x1 - 9y/y1 = 7... ==> get equation of ellipse, then find eccentricity.

one small thing...r u sure M = (7x1/32, -7x1/18)?

 

[c0okies]

oh oops sorry i meant M= 7x1/32, -7y1/18

thanx

 

[c0okies]

i tried subbing it in but all i end up with is 7=7 at the end.. which isnt the answer Dx

 

[Mountain.Dew]

i tried subbing it in but all i end up with is 7=7 at the end.. which isnt the answer Dx

okay i think i know what to do now...

using x=7x1/32 and y = -7y1/18, use the fact that x1^2/16 + y1^2/9 = 1

do simultaenous equations, eliminating x1's and y1's to get an equation in terms of x's and y's

that should do the trick! *fingers crossed*

 

[c0okies]

LoL! well i tried it before and it didnt work

i even transformed it to a cartesian equation and it didnt work out! *cries

 

[Mountain.Dew]

LoL! well i tried it before and it didnt work

i even transformed it to a cartesian equation and it didnt work out! *cries

please, dont be disheartened. when theres a will, theres a way!

try this:

x=7x1/32 ....(1)
y= -7y1/18 ...(2)
x1^2/16 + y1^2/9 = 1...(3)

now, (1) --> x^2 = 49x1^2/1024 --> x1^2 = 1024x^2/49

so x1^2/16 = 1024x^2/(49*16) = 64/49x^2

same procedure with (2) --> eventually u get y1^2 / 9 = 36/49y^2

so, substituting (1) and (2) into (3), we get ==> [64/49x^2] + [36/49y^2] = 1 <== ur locus equation

now then co0kies, can u go from there?

 

[c0okies]

omG now i get it xD

yes i can from there; thanku!

 

[Mountain.Dew]

omG now i get it xD

yes i can from there; thanku!

*throws around a 2 cent coin*

hehe a pleasure co0kies, a pleasure *bows*

 

[c0okies]

xD