Conics ....help (1 Viewer)

[haboozin]

Conics Help Marathon

<b>QUESTION 1:</b>

P(acos@,bsin@) lies on the ellipse x²/a² + y²/b² = 1, where a > b > 0. The tanget to the ellipse at P passes through a focus of the hyperbola x²/a² - y²/b² = 1 with eccentricity e.


i, Show that the tangent to the ellipse at P has equation bxcos@ + aysin@ = ab

(i think thats a typo cos i get bxcos@ + aysin@ = ae)

ii,
show that P lies on a directrix of the hyperbola.


iii,
show that the tangent to the ellipse at P has gradient +-1


<b>QUESTION2:</b>

ok i know this has something to do with relationships, but i forgot these relationships..




P(acosA,bsonA) and Q(acosB,bsinB) lie on the ellipse x²/a² + y²/b² = 1

i,
verify that the coordinates of P satisfy x/a cos((A + B)/2) + y/b sin((A + B)/2) = cos((A - B)/2)

Note: A and B are angles and have nothing to do with a and b.

ii, deduce that the chord PQ has equation x/a cos((A + B)/2) + y/b sin((A + B)/2) = cos((A - B)/2)

this is very easy really, because if p works in there so willl q therefore it willl be a chord..

iii, hence or otherwise show that if PQ is a focal chord of the ellipse, then cos((A-B)/2) = +-ecos((A + B)/2)


<b>QUESTION3:</b>

The line l is a common tangent to the hyperbolas xy=cb², xb²/ab² - yb²/bb²=1 with points of contact P (ct, c/t) and Q (asec@,btan@) respectively

you may assume: tangent at P is x +t²=2ct and tangent at Q is xsec@/a - ytan@/b = 1

deduce that:
sec@/a = -tan@/bt^2 = 1/2ct




<b>QUESTION 4:</b>

P (asec@,btan@) lies on the hyperbola x²/a² - y²/b² = 1. the tangent at P is also a tangent of a circle. (x - ae)² + y² = a²(e²+1)


show that sec@ = -e

Ok, for this question first i tried to solve the equation of the tangent of the hyperbola simultaniously with the circle when DELTA = 0. That came out to be too long! so then i tried to implicietly differentiate both the circle and the hyperbola then put in P in both of then and solve them when they both equal and i got sec@ = 1/e . I then tried to use how tangents to the circle are always perpendicular to the center and still got sec@ = 1/e

btw, this was in our term test today.




If you are doing these, full working will be appreciated dont just tell me what to do because i already have at least some idea of how to do these questions and i really need to see the proofs.

 

[shafqat]

P(acos@,bsin@) lies on the ellipse x²/a² + y²/b² = 1, where a > b > 0. The tanget to the ellipse at P passes through a focus of the hyperbola x²/a² - y²/b² = 1 with eccentricity e.


i, Show that the tangent to the ellipse at P has equation bxcos@ + aysin@ = ab

(i think thats a typo cos i get bxcos@ + aysin@ = ae)

ii,
show that P lies on a directrix of the hyperbola.


iii,
show that the tangent to the ellipse at P has gradient +-1

i the tangent does have equation bxcos@ + aysin@ = ab
differentiate: dy/dx = -bcos@/asin@
then use point gradient formula, multiply out and simply

ii. sub in (ae,0): so bcos@ae = ab
cos@ = 1/e
So x-coordinate of P is a/e, so P lies on the directrix of the hyperbola

iii) We know cos@ = 1/e, draw a triangle and work out the gradient, namely -bcos@/asin@

 

[haboozin]

i the tangent does have equation bxcos@ + aysin@ = ab
differentiate: dy/dx = -bcos@/asin@
then use point gradient formula, multiply out and simply

i get

bxcos@ + aysin@ = abecos@

how do u prove ecos@ = 1
(if u can prove this ii will b done in 1 step)

 

[haboozin]

edited out, its in the original post now..

 

[shafqat]

i get

bxcos@ + aysin@ = abecos@

how do u prove yasin@ + xbcos@
(if u can prove this ii will b done in 1 step)

differentiate parametrically to get the gradient:
dy/dx = -bcos@/asin@
then use pt gradient formula:
y - bsin@ = -bcos@/asin@ (x - acos@)
asin@y - absin^2@ = -bcos@x + abcos^2@
yasin@ + xbcos@ = ab(cos^2@ + sin^2@)
yasin@ + xbcos@ = ab

ii. sub (ae,0) into the equation:
bcos@ae = ab
cos@ = 1/e

 

[haboozin]

shafqat,

any chance of you asnwering my other question??

 

[KFunk]

NEXT QUESTION:

ok i know this has something to do with relationships, but i forgot these relationships..

P(acosA,bsonA) and Q(acosB,bsinB) lie on the ellipse x₂/a₂ + y₂/b₂ = 1

i,
verify that the coordinates of P satisfy x/a cos((A + B)/2) + y/b sin((A + B)/2) = cos((A - B)/2)

You could just sub P(acosA, bsinA) into the equation and do some fancy algebra. Alternatively you could show that the equation is that which describes the chord PQ (which it is). For the latter method the biggest hurdle is finding the gradient in a suitable form then it kinda falls into place:

b(sinA - sinB)/a(cosA - cosB)

= b[2sin((A - B)/2)cos((A+B)/2)] / a[-2sin((A-B)/2)sin((A+B)/2)]

= -bcos[(A+B)/2] / asin[(A+B)/2]


Use that along with coordinate P and you should get your equation.

The other slightly weird bit is when you get cosAcos[(A+B)/2] + sinAsin[(A+B)/2] but just do the product/sum/difference stuff to get cos[(A-B)/2]

Perhaps subbing the point in would be easier? I dunno... You try

 

[haboozin]

The other slightly weird bit is when you get cosAcos[(A+B)/2] + sinAsin[(A+B)/2] but just do the product/sum/difference stuff to get cos[(A-B)/2]

Perhaps subbing the point in would be easier? I dunno... You try

OH INDEED YEA stupid me, i see this now...


ok so,
i and ii answered

how bout iii and Q3... ?

 

[shafqat]

For question 3, the two different equations are one and the same, as it is a common tangent.
Equation 1: xsec@/a - ytan@/b = 1
Equation 2: x + t^2y = 2ct
Dividing by 2ct, x/2ct + yt/2c = 1

This is the important step: equate coefficients of x and y, as the equations are exactly the same.
So equating coeff of x, sec@/a = 1/2ct
So equating coeff of y, -tan@/b = t/2c

 

[haboozin]

This is the important step: equate coefficients of x and y, as the equations are exactly the same.
So equating coeff of x, sec@/a = 1/2ct
So equating coeff of y, -tan@/b = t/2c

wow, you make the best decisions...
no wonder you came what... 5th in the state...

 

[haboozin]

<b>Please try <i>QUESTION 4</i> for me guys</b>

 

[haboozin]

shafqat Q4 please...

 

[FinalFantasy]

u might wana just make a new post.. some ppl dun look in posts w\ many posts in it already