MedVision ad

Revision of locus and parabola + Sequence and series (1 Viewer)

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
Heya this is a thread where anyone can post difficult, regular or extremely challenging questions from the locus, parabola, sequence and series topic. This thread was started because i have a test on friday and hopefully i can revise on 2u topics which i did last year xDD
Thanks and much appreciated
Very easy general questions are acceptable too :eek:
Free rep up for grabs, just don't scab them all unless the questions are left unanswered.
Someone start it :cool:
 

hscishard

Active Member
Joined
Aug 4, 2009
Messages
2,033
Location
study room...maybe
Gender
Male
HSC
2011
Show that the locus of a point, which moves so as always to be three times further from one fixed point than from another fixed point, is a circle.

1 year spent trying to show that. Still no sucess
 

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
Show that the locus of a point, which moves so as always to be three times further from one fixed point than from another fixed point, is a circle.

1 year spent trying to show that. Still no sucess
simply state it is a circle with radius 3 i guess

now my question
prove that the line 3x+4y+21 = 0 is a tangent to the circle x² - 8x + y² +4y-5 = 0
should not be hard
 

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
clue : use perp distance formula to prove it has same distance as radius
 

hscishard

Active Member
Joined
Aug 4, 2009
Messages
2,033
Location
study room...maybe
Gender
Male
HSC
2011
You sure? It doesn't sound right

As for your question. Too long to type up. Just simply complete the sqaure to get it to general circle form. Then make x or y the subject on the straight line. Sub the points into x or y. Simplify(good luck typer). Then sub this point back to circle(i think not straight line as circles have a sqaured in it)

Should be one point. Thats how I would do it. Is there anoher way?
 

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
|ax + bx + c| / (root a^2 + b^2) ------> perp distance form
sub in a b c relatively and x as the centre of the circle, thus obtaining a distance which = 5.
Since the circle has radius = 5, it is therefore a tangent.
As for your method, it is extremely tedious to make LHS = RHS
Nevertheless i dont even think you can do this, prove me wrong if you can
 
Last edited:

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
Moving on to the next question
A point P(x, y ) moves in the XY - plane such that its perp distance to the line 2x - y +5 = 0 is equal to its perp distance to the line 2x + 4y - 7 = 0. Find the path of P and describe how it is related to the given lines.

Locus questions preferred over sequence
Post questions please :p
 

super.muppy

Member
Joined
Aug 29, 2008
Messages
375
Gender
Male
HSC
2010
|ax + bx + c| / (root a^2 + b^2) ------> perp distance form
sub in a b c relatively and x as the centre of the circle, thus obtaining a distance which = 5.
Since the circle has radius = 5, it is therefore a tangent.
As for your method, it is extremely tedious to make LHS = RHS
Nevertheless i dont even think you can do this, prove me wrong if you can
why not just prove that it touches the circle at one point by simultaneous since the basic definition of a tangent is a line that touches a curve at one point. saves time
 
Last edited:

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
<DIR>find the exact length of the tangent from (4, - 5) to the circle x^2 + 4x + y^2 - 2y - 11 = 0
</DIR>
next
 

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
YAYER SOLVED IT MYSELF LOL
now changing the question around
how do you find the point where the tangent touches the circle
now thats idno xD
 

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
why not just prove that it touches the circle at one point by simultaneous since the basic definition of a tangent is a line that touches a curve at one point. saves time
your question was answered in my quote :p
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
YAYER SOLVED IT MYSELF LOL
now changing the question around
how do you find the point where the tangent touches the circle
now thats idno xD
that's easy. differenciate implicitly, then let that be equal to the gradient of the tangent. Form a relationship between x and y.

Sub that relationship into the circle, then solve for x and y. you will get 2 solutions of x and y (since there are 2 places on a circle that can have tangents with the same gradient). Reject the incorrect one using your own logic and reasoning.

simple question.
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Moving on to the next question
A point P(x, y ) moves in the XY - plane such that its perp distance to the line 2x - y +5 = 0 is equal to its perp distance to the line 2x + 4y - 7 = 0. Find the path of P and describe how it is related to the given lines.

Locus questions preferred over sequence
Post questions please :p
By looking at the question, it appears that the locus is the line which bisects the angle made from these two lines intersecting. As to finding it, I would use the angle between two lines formula to find the gradient, and then use the point of intersection of the other two lines.
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Show that the locus of a point, which moves so as always to be three times further from one fixed point than from another fixed point, is a circle.

1 year spent trying to show that. Still no sucess
to do this question, let the two points be (a,0) and (b,0) because it makes it easier to do the question on the x axes, though any will work. Let the point of the locus be P, such that PA = 3PB

Use the distance formula so you get (x-a)^3 + y^2 = 3 [ (x-b)^2 + y^2 ]

Simplify to get x^2 + y^2 + (a-3b)x = (a^2-3b^2)/2 which is obviously a circle.

challenging to figure out the first step, but simple once you get it.

simply state it is a circle with radius 3
sorry, but that's wrong. 3 times the distance from a fixed point is not the same as the distance being 3 units.
 

cutemouse

Account Closed
Joined
Apr 23, 2007
Messages
2,250
Gender
Undisclosed
HSC
N/A
Simplify to get x^2 + y^2 + (a-3b)x = (a^2-3b^2)/2 which is obviously a circle.
You need to get it in the form (x-h)^2 + (y-k)^2 = r^2 to show that it is a circle... otherwise it could be an ellipse or another conic section.
 

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
that's easy. differenciate implicitly, then let that be equal to the gradient of the tangent. Form a relationship between x and y.

Sub that relationship into the circle, then solve for x and y. you will get 2 solutions of x and y (since there are 2 places on a circle that can have tangents with the same gradient). Reject the incorrect one using your own logic and reasoning.

simple question.
its easy to say but can you do it?
next question
<DIR>Find the equation of the focal chord that cuts the curve y^2 =16x at (4,8)
its x = 4 but dno how to prove it even though you can see they are both at x = 4
</DIR>
 

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
dw ill just take the easy way out and say x= 4 cos both points are vertical
next question please
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top