Conics Questions (help!) (1 Viewer)

[Trev]

Here's two questions, have fun (I don't know how to do them....)

Both from Fitzpatrick (New Senior) Exercise 32(c).

8) The equations of the two ellipses are (i) 4x² + 9y² = 36, (ii) 2x² + 3y² = 30. A tangent to ellipse (i) meets the ellipse (ii) at the points P and Q. Show that the tangets at P and Q to ellipse (ii) are at right angles to one another.

We can do this question by geometry, however not by algebra. Can anyone do it? Or if not explain why it will not work?

9) P is a variable point on the ellipse with equation x²/a² + y²/b² = 1 and S and S' are the foci. Show that PS and PS' are equally inclined to the tangent at P.

Please attempt, thanks!

 

[Slidey]

8) The equations of the two ellipses are (i) 4x² + 9y² = 36, (ii) 2x² + 3y² = 30. A tangent to ellipse (i) meets the ellipse (ii) at the points P and Q. Show that the tangets at P and Q to ellipse (ii) are at right angles to one another.

You can do it algabraicly. I know that the two tangents meet at an exterrnal point which MIGHT suggest chord of contact stuff... but. An algebra bash is not pretty. I might be of help tomorrow when I've revised some conics.

The algebra bash: get the eqn of tangent to ellipse (i), solve simultaneously in ellipse (ii), get points of intersection, get gradient at those points, prove both gradients multiplied = -1

 

[KFunk]

I really wish that conics would curl up in a hole and die.

 

[Trev]

Slide, my teacher did this question but he ended up with something (I forget) along the lines of "sinx - e" having to equal "sinx + e", inside of the tanx=|{m1-m2}/{1 + m1m2}| formula, which cannot be true unless e=0, which is only true for circles.... (I think, correct me if I am wrong!).

Anyone up for a try?

 

[Slidey]

Trev, do what I said. I did not use angle between two lines (why bother?).

 

[Trev]

Mmmk, i'll do what you said.... I didn't bother reading your post properly before

 

[DistantCube]

Question 8 is a ripper, so is question 9 for that matter...but I'll give you the jist;


Q8:

edit: I implicitly differentiated to get the gradients and used the line formula to get PR and QR...

let R be (x0, y0)
P be (x1, y1)
Q be (x2, y2)

PR is 2xx1 + 3yy1 = 30 ----(1)
QR is 2xx2 + 3yy2 = 30 ----(2)
R is 2x^2 + 3y^2 = 30 ----(3)
and so has equation;
2xx0 + 3yy0 = 30 ----(4)
Since this chord touches 4x^2+9y^2 = 36 -------(5)
then when we solve (4) and (5) simultaneously we get equal roots, i.e: discriminant = 0.

So 4x^2 + 9(30-2xx0/all over/3y0)^2 = 36
...
4x^2(x0^2 + y0^2) - 120xx0 + 36(25-y0^2) = 0
discriminant = 900x0^2 - 900(x0^2+y0^2) + 36(x0^2+y0^2)y0^2
=36y0^2(x0^2 + y0^2 - 25)
=0
Hence locus of R is x0^2 + y0^2 = 5, a circle -----(6)

now we want to show PQ subtends a right angle at R, i.e: mPR . mQR = -1
i.e: (-2x1/3y1)(-2x2/3y2) = -1 using (1) and (2)
OR
4x1x2 + 9y1y2 = 0 (**)

if we solve the equations (4) and (3), the solutions for x will be the values x1 and x2

2xx0 + 3yy0 = 30 -----(4)
2x^2 + 3y^2 = 30 ----(3)

So; 2x^2 + 3(30-2xx0/all over/3y0) = 30
work that through we get;
x^2(2x0^2+3y0^2)-60xx0 + 45(10-y0^2) = 0
(we use x0^2 + y0^2 = 25)
x^2(2x0^2 + 3(25-x0^2))-60xx0 + 45(10-25+x0^2) = 0
x^2(75-x0^2) - 60xx0 + 45(x0^2-15) =0
therefore, x1 + x2 = 60x0/(75-x0^2)
and x1x2 = 45(x0^2-15) /all over/ (75-x0^2)

Now back to (**) which is the thing we are trying to show:
4x1x2+9y1y2
=4x45(x0^2-15)/all over/75-x0^2) + 9((30-2x0x1)/3y0)((30-2x0x2)/3y0)
which eventually works out to be:
7200x0^2 - 180x0^4 - 67500 + 67500 - 7200x0^2 + 180x0^4 /all over/ (75-xo^4)(25-x0^2)

=0

but there must be an easier way, that's really messy...I tried it again and it came out messy again, so I just left it, it works anyway.

I'll post Q9 in a mo.

 

[Trev]

Dang, thanks DistantCube; i'll have a look through it now!

 

[DistantCube]

Ok now Question 9;
(I always use "//" as "all over")

angle between two lines; tan @ = | m1-m2 // 1+m1m2 |

at P:
dy/dx = dy/d@ . d@/dx
= b cos @ . 1/-asin@
m1= -b/a . cos@/sin@

at PS:
m2 = bsin@-0 // acos@-ae

at PS' : m3 = bsin@ // acos@+ae

therefore angle between tangent and PS:

tan a = | [-bcos@/asin@ - bsin@/a(cos@-e)] // [1 + -bcos@/asin@ . bsin@/acos@-e] |

work that through, cancelling etc, you end up with

-ab+abecos@ // [sin@(a^2(cos@+e)-b^2cos@]

Now, similarly;

tan b = | [-bcos@/asin@ - bsin@/(acos@+e)] // [1 + -bcos@/asin@ . bsin@/a(cos@+e)] |

working through and cancelling again, you end up with

| [-ab(cos^2@+sin^2@)-abecos@] // [sin@(a^2(cos@+e)-b^2cos@)]

we need to show they're the same, so we need: a=b

b^2=a^2(1-e^2)
a^2b^2=a^2e^2

for tan a;

= | ab(cos@-1) // sin@(a^2cos@-b^2cos@-a^2e)

which ends up as:

tan a = b/aesin@

and for tan b;

tan b = | [-ab-aecos@] // [sin@(a^2(cos@+e)-b^2cos@]

= | ab(1+ecos@) // [sin@(cos@(a^2e^2)+a^2e] |

<<< where did the negatives on the top line go? It's absolute value, we have two negatives, therefore its positive >>>

work that down we end up with, surprise surprise;

tan b = b/aesin@
=tan a

done.

 

[Trev]

Thanks DistantCube and Slide!

 

[Slidey]

Nobody has found an easy way?

 

[Trev]

It's up to you Slide. Maybe,
Templar? The other guy with the kool avatar? ngai? KeyPad?.....

 

[Captain pi]

(8)

I tried doing this, but ended giving up in disgust.

Every method I tried required knowledge of the coordinates of the intersection of the tangent (of the smaller ellipse) to the larger ellipse. The coordinates seemed mightily convoluted.

I cannot remember seeing a question of such difficulty (except Q8 in 1989) in a 4U paper.

 

[Trev]

Yeah Hugh, Mr Di Bona said that you attempted and got as far as he did, he said you're more intelligent than him lol.
Hmm 1989... I hope they don't feel the need for this year to be another difficult question

 

[Trev]

DistantCube; how long did it take you to complete each question?

 

[Captain pi]

Yeah Hugh, Mr Di Bona said that you attempted and got as far as he did, he said you're more intelligent than him lol.
Hmm 1989... I hope they don't feel the need for this year to be another difficult question

I think this year will be a more difficult year, but I think they'll stick to topics you have learnt. (I think [correct me if I am wrong, Olympians et al.] that that question was on Number Theory; it was pretty juicy, but quite difficult in an exam situation.)

 

[DistantCube]

DistantCube; how long did it take you to complete each question?

oh a fair while, Question 9 probably 15-30 mins, and Question 8 took me a touch over an hour because I thought I was doing it wrong because it looked so damn messy, but eventually I kept getting the same thing, parametrically and cartesian, so I just carried on and it turned out to work, as ugly as it was, gave my teacher a go at question 8 too, came up with the same as me, so yeah, he said there should be a neater way but he doesn't know.

 

[ngai]

8) The equations of the two ellipses are (i) 4x² + 9y² = 36, (ii) 2x² + 3y² = 30. A tangent to ellipse (i) meets the ellipse (ii) at the points P and Q. Show that the tangets at P and Q to ellipse (ii) are at right angles to one another.

this method looks long...too long...but anyway
let those tangents meet at T, and let the tangent to (i) meet (i) at R
so PT is 2xx_P+3yy_P=30, giving m_PT=(-2x_P)/(3y_P)
and QT is 2xx_Q+3yy_Q=30, giving m_QT=(-2x_Q)/(3y_Q)
we want to prove that those two gradients multiplied gives -1
which is the same as proving:
4x_Px_Q=-9y_Py_Q

ok, now P and Q are intersections of the tangent at R with the 2nd ellipse
tangent at R is: 4xx_R+9yy_R=36, which can be rearranged to get:
(1) x = 9(4-yy_R)/(4x_T)
(2) y = 4(9-xx_R)/(9y_T)
sub (1) into 2x²+3y²=30, expand and simplify to get a quadratic in y
this quadratic has roots y_P and y_Q, so get the value of y_Py_Q by relationship between roots and coefficients:
y_Py_Q = 8(162-30x_R²)/(24x_R²+81y_R²)
similarly, by sub (2) into 2x²+3y²=30,
x_Px_Q = 27(48-30y_R²)/(54y_R²+16x_R²)
now, in both of those, replace y_R² with (36-4x_R²)/9, since R lies on ellipse (i)
then expand and simplify, and you should get 4x_Px_Q=-9y_Py_Q

hope i havent mistyped anything

 

[ngai]

9) P is a variable point on the ellipse with equation x²/a² + y²/b² = 1 and S and S' are the foci. Show that PS and PS' are equally inclined to the tangent at P.

There is a theorem that states:
In a triangle ABC, with D on BC, AD bisects angle A if and only if AB/AC = BD/DC

so in our ellipse situation:
P(acos@,bsin@), and let N be the x-intercept of the normal at P,
then we want to show that PS'/PS = NS'/NS
(which would mean S'PN = SPN, and so tangent is equally inclined to PS and PS')

PS' = ePM' and PS=ePM, so:
PS'/PS
= PM'/PM
= |acos@ - (-a/e)|/|acos@ - a/e|
= |1+ecos@|/|1-ecos@| (1)

now, normal at P is ax/cos@ - by/sin@ = a²e² (note: a²e² = a² - b²)
so to get coords of N, y = 0:
ax/cos@ = a²e²
therefore x_N = ae²cos@
therefore NS'/NS = |ae²cos@ - (-ae)|/|ae²cos@ - ae|
= |ecos@ + 1|/|ecos@ - 1| (2)

so by (1) and (2), you get the result

 

[Trev]

Ah, thanks for that ngai! Yes, alot quicker than DistantCube's, took me ages to write that out!