Parametrics: Chord of Contact (2 Viewers)

[namburger]

Just had my 3u exam and a question involving chord of contact came up in which i just quoted the formula.
Can someone explain to me why the chord of contact has the same formula of the tangent. I can't understand it logically.

 

[tacogym27101990]

it doesnt.
tangent has equation: y-px+ap^2 = 0
chord of contact has: xx* =2a(y + y*)
where (x*,y*) is the intersection of the tangents

 

[namburger]

it doesnt.
tangent has equation: y-px+ap^2 = 0
chord of contact has: xx* =2a(y + y*)
where (x*,y*) is the intersection of the tangents

It does, you're using PARAMETERS
if you use x1,y1 you will get the same equation for the tangent and the chord of contact

 

[tacogym27101990]

sorry i just figured u were using parameters from the heading..

 

[doink]

perhaps tried deriving it?

x^2=4ay

x=2at
y=at^2

y=x^2/4a
y'=x/2a
=t (at x=2at)

y-y1 = m(x-x1)
y-at^2 = t(x-2at)
y-at^2 = tx - 2at^2
y-tx+at^2 = equation of tangent

chord on parabola = y=(p+q)x/2 - apq

you should get it from there

 

[namburger]

perhaps tried deriving it?

x^2=4ay

x=2at
y=at^2

y=x^2/4a
y'=x/2a
=t (at x=2at)

y-y1 = m(x-x1)
y-at^2 = t(x-2at)
y-at^2 = tx - 2at^2
y-tx+at^2 = equation of tangent

chord on parabola = y=(p+q)x/2 - apq

you should get it from there

1. I'm not using parameters, im using x1,y1
2. I know how to derive it but don't understand it. Its like the tangent at point P(x1,y1) has tangent of equation xx1=2a(y+y1). The tangent at point Q(x2,y2) has tangent of equation xx2=2a(y+y2). These 2 tangents intersect at (x0,y0) so therefore the chord joining P and Q must have equation xx0=2a(y+y0)??

 

[conics2008]

tangent = xx1=2a(y+y1) is not the same as xx1=2a(y+y1)

that 2nd formula states the chord of contact ....

if i give you an external point and a parabola x^2=4ay with point (x1,y1) and askd you to find chord of contact you would have to use that formula you stated other wise it would be imposible ..

dont get confused just because they have the same equation doesn't mean they are the same =) peace..

 

[dood09]

yep

 

[undalay]

tangent = xx1=2a(y+y1) is not the same as xx1=2a(y+y1)

that 2nd formula states the chord of contact ....

if i give you an external point and a parabola x^2=4ay with point (x1,y1) and askd you to find chord of contact you would have to use that formula you stated other wise it would be imposible ..

dont get confused just because they have the same equation doesn't mean they are the same =) peace..

He's asking why the equation is the same [xx1=2a(y+y1)] even though they are different [tangent / chord of contact] you dck. L2comprehend.

 

[undalay]

xx1 = 2a(y+y1) ... (1)
xx2 = 2a(y+y2) ... (2)

Let T(x0,y0) be external point and A(x1,y1), B(x2,y2) two points on parabola).

(1) and (2) are equations of tangents.

(1)...
T must lie on tangent (1)
x0x1=2a(y0+y1)
THUS A must lie on the line.
x0x = 2a(y0 + y) ...(3)

(2)...
T must lie on tangent (2)
x0x2=2a(y0+y2)
THUS B must lie on the line.
x0x = 2a(y0 + y) ...(4)

BUT (3) = (4)
x0x = 2a(y0 + y) holds both point A and B.

Therefore x0x = 2a(y0 + y) is the chord of contact.

 

[namburger]

xx1 = 2a(y+y1) ... (1)
xx2 = 2a(y+y2) ... (2)

Let T(x0,y0) be external point and A(x1,y1), B(x2,y2) two points on parabola).

(1) and (2) are equations of tangents.

(1)...
T must lie on tangent (1)
x0x1=2a(y0+y1)
THUS A must lie on the line.
x0x = 2a(y0 + y) ...(3)

(2)...
T must lie on tangent (2)
x0x2=2a(y0+y2)
THUS B must lie on the line.
x0x = 2a(y0 + y) ...(4)

BUT (3) = (4)
x0x = 2a(y0 + y) holds both point A and B.

Therefore x0x = 2a(y0 + y) is the chord of contact.

OK buddy =]

 

[advanced sam]

sorry i just figured u were using parameters from the heading..

der, :haha

god i hate parameters, not gonna help you here ill end up being wrong