Conics!!! (1 Viewer)

[zeek]

okay i have no idea what i am doing wrong but i cannot for the life of me get this answer right even though i'm pretty sure everything is right.
Find the locus of a variable point P(x,y) which moves so that its distance from (1,0) is one third its distance from x=9.

Don't you let PS=e.(1/3).PM where PS is the distance from the focus and PM is the distance to the directrix?

 

[BIRUNI]

Do you get 2/3 x^2+y^2=2

 

[onebytwo]

okay i have no idea what i am doing wrong but i cannot for the life of me get this answer right even though i'm pretty sure everything is right.
Find the locus of a variable point P(x,y) which moves so that its distance from (1,0) is one third its distance from x=9.

Don't you let PS=e.(1/3).PM where PS is the distance from the focus and PM is the distance to the directrix?

why do you need the e?
i thought its just one of those basic 2 unit locus questions
i got an ellipse, equation 8x^2 + 9y^2 = 72

 

[zeek]

Do you get 2/3 x^2+y^2=2

Nah i get 80x²/81 - 16x/9 + y²=0 but the answer is x²/9 + y²/8=1

why do you need the e?
i thought its just one of those basic 2 unit locus questions

No, this is for the equation of an ellipse so it has an eccentricity 0<e<1...

 

[onebytwo]

oh sorry, you mean the e is the one-third,
silly me

 

[onebytwo]

PS = 3PM
so PM^2 = 9PS^2
ie 9(y^2 + x^2 -2x+1) = x^2 - 18x+81
then simplify to get 8x^2 + 9y^2 = 72
then divide through by 72

 

[zeek]

ok ok OMG i got it but this time i left out the e... I thought you had to put e in the equation?

 

[NickP101]

No, e is the ratio of PS/PM (distance from point to focus/distance from point to directrix). It says PS = PM/3, so PS/PM = 1/3 which = e.

 

[zeek]

Okay can someone please do this question...

P(acos @, bsin@) and Q(acos %, bsin%) lie on an ellipse. Show that if PQ subtends a right anle at (a,0) then tan(@/2)tan(%/2)=-b²/2².

I know what i have to do but i just can't do it... if that makes sense

 

[onebytwo]

Okay can someone please do this question...

P(acos @, bsin@) and Q(acos %, bsin%) lie on an ellipse. Show that if PQ subtends a right anle at (a,0) then tan(@/2)tan(%/2)=-b²/2².

I know what i have to do but i just can't do it... if that makes sense

i trust you mean = - b^2/a^2

find the gradient of PS and QS where S is (a,0) and since they subtend a right angle they multiple to get -1.
so (b^2sin@sin%)/(a^2(cos@-1)(cos%-1)) = -1
break down the sin @ and the sin % to half angles in the numerator ie 2sin(@/2)cos(@/2) and 2sin(%/2)cos(%/2) respectivley
break down the cos@ and the cos% to 1-2sin^2@ and 1-2sin^2% respectively
to get to (b^2cos(@/2)cos(%/2))/(a^2sin(@/2)sin(%/2)=-1
simplifly in terms of tan(@/2) and tan(%/2) to the required relationship
EDIT:too slow

 

[zeek]

thx guys

 

[zeek]

Ok here's another one lol

P(asec@,btan@) lies on the hyperbola. The tangent at P cuts the x-axis at X and the y-axis at Y. Show that PX/PY = sin²@ and deduce that if P is an extremity of a latus rectum, then PX/PY = (e²-1)/e²

I've done the first bit, but for the second bit, do all i have to do is sub ae for the x value of P rather than asec@ because it's on the focal chord?