Conics question help (1 Viewer)

[king.rafa]

P (sec x, b tan x) lies on the hyperbola x^2/a^2 - y^2/b^2 = 1. The tangent at P cuts the x-axis at X and the y-axis at Y. Show that PX/PY = sin^2 x and deduce that if P is an extremity of a latus rectum, then PX/PY = (e^2-1)/e^2.

 

[kooltrainer]

dun u jus find the intersection, distance formula. then let x = ae

 

[undalay]

i challenge conics2008 to do this question.

$5 i can give a better proof.

 

[ashraf91]

instead ov being a dic y dont u jst help

 

[kony]

try not using the distance formula.

 

[Slidey]

i challenge conics2008 to do this question.

$5 i can give a better proof.

Seconded.

 

[namburger]

i challenge conics2008 to do this question.

$5 i can give a better proof.

why challenge haha

 

[conics2008]

i challenge conics2008 to do this question.

$5 i can give a better proof.

OK..

Equation of Tangent = xx1/a^2 - yy1/b^2 = 1

xsec(x)/a - ytan(x)/b =1

at y=0 x = a/sec(x) >> X ( a/sec(x),0)
at x=0 y= -b/tan(x) >> Y( 0,-b/tan(x))

USING SIMILAR TRIANGLES PMX AND PNX

px/py = pm/pn = btan(x)/btan(x)+b/tan(x)
= btan^2 (x)/btan^2 (x) +b
= tan^2 (x)/ sec^2 (x)
= sin^2 (x)

there you go thats the first part...

if p is an extremity of a latus rectume ( wow sounds sexual ) it must pass through (ae,0) right ?:worried:

then asec(x) = ae >>> sec(x) = e

now from px/py = sin^2 (x)
= 1-cos^2 (x)
= 1 - 1/e^2 >>> e^2-1/e^2


Good day kids now give me my 5 bucks lol....



EDIT: Sorry I'm abit late but I've done this question already, this is the shortest method I know.. =)

 

[undalay]

2 late conics2008 ):

People already posted up the solutions.

Let's play again next time :]

 

[conics2008]

Undalay, I like you lol... I still want my 10 bux !


PS: I'm not gay =)

 

[conics2008]

lol.. you're look at you're solution and look at mine, you coudn't even do full working out. This is how I solve this problem...

tangent equation: xsec@/a - ytan@/b = 1
Y coordinates: (0, - bcot@)

by similar triangles: PX/PY = btan@ / (btan@ + bcot@)
= tan@ / [ (tan^2@ + 1) / tan@ ]
= tan^2@ / (1 / cos^2@)

This is what you wrote look at mine.. think before you talk .... i dont need you're crappy working out =) HOMO

 

[kooltrainer]

relax.. jus conics

 

[conics2008]

ok if im a homo then you got beatin by a homo... you're sad seriously.. get a life troll.