4U Maths Question Thread

[cmadeleine]

For now, I am going to redirect you to the discussion here.

Use the resources I have linked to.

Thank you so much!!

[kiwiberry]

Hey could please I get some help on these

1) P(cp,c/p) and Q(-cp,-c/p) are points on the rectangular hyperbola xy=c². The normal at P to the hyperbola meets the curve again at R. Find the coordinates of R, and hence show that \(\angle PQR\) is a right angle

2) P(3p,3/p) and Q(3q,3/q) are points on different branches of the hyperbola xy=9. The tangents at P and Q meet at T. Find the locus of T if the positions of P and Q vary so that the chord PQ passes through (0,4).
I managed to find the locus as x=9/2, but the answer says that it is x=9/2, y<0. I was wondering why y has to be less than 0? T is \(( \frac{6pq}{p+q} ,\frac{6}{p+q})\) for reference

Thank you!

[jakesilove]

Hey could please I get some help on these

1) P(cp,c/p) and Q(-cp,-c/p) are points on the rectangular hyperbola xy=c². The normal at P to the hyperbola meets the curve again at R. Find the coordinates of R, and hence show that \(\angle PQR\) is a right angle

2) P(3p,3/p) and Q(3q,3/q) are points on different branches of the hyperbola xy=9. The tangents at P and Q meet at T. Find the locus of T if the positions of P and Q vary so that the chord PQ passes through (0,4).
I managed to find the locus as x=9/2, but the answer says that it is x=9/2, y<0. I was wondering why y has to be less than 0? T is \(( \frac{6pq}{p+q} ,\frac{6}{p+q})\) for reference

Thank you!

Hey!

Let's find the normal at P


So, the normal will be






So,



This meets the hyperbola again where y=c^2/x



Solve this, and you'll get



The first one is obvious (it's the point from which the normal is from!), but the second one is what we're looking for.

The y value will be



Now, you just need to find the equation of the lines from P to Q, and Q to R. Show that the two gradients are inverse negatives of each other, and bam! Q1 done.

I'll work on Q2 now.


Okay, so for Q2, I'm actually not sure! Might have to call in the infinite wisdom of RuiAce here

[beau77bro]

hey jake, or whoever decides to answer this. im having trouble with this question and its kinda put me off the whole of fitzpatrick ahahha. but i was just wondering how we are supposed to approach this? question 6b

[kiwiberry]

Hey!

Let's find the normal at P


So, the normal will be






So,



This meets the hyperbola again where y=c^2/x



Solve this, and you'll get



The first one is obvious (it's the point from which the normal is from!), but the second one is what we're looking for.

The y value will be



Now, you just need to find the equation of the lines from P to Q, and Q to R. Show that the two gradients are inverse negatives of each other, and bam! Q1 done.

I'll work on Q2 now.


Okay, so for Q2, I'm actually not sure! Might have to call in the infinite wisdom of RuiAce here

Thank you :-)

[Shadowxo]

hey jake, or whoever decides to answer this. im having trouble with this question and its kinda put me off the whole of fitzpatrick ahahha. but i was just wondering how we are supposed to approach this? question 6b

(Image removed from quote.)

Just expand it out and go from there
x + xi + 2y - 3yi = 10
(x + 2y) + (x - 3y)i = 10
x + 2y = 10
x - 3y = 0
5y = 10
y = 2
x = 6

[beau77bro]

hello again i have a lot of questions about section 3.2 of 4u cambridge, mostly the parts involving multiple parametric points on ellipses or hyperbola. so the parts with chords between points: P (acosß, bsinß) and Q (acos∂, bsin∂). firstly, in the explanation/ working to show the lines equation and gradient there are a lot of steps missing and i cant understand how they get anywhere, mostly the gradients? secondly, how relevant/do we need to know this at all because my teacher doesnt even know how to do it really i dont think? and if we need to know, is it at a level of deriving, im gonna assume so because it cant hurt but still?

so yea if anyone wants to shoot up some kind of working for the gradients that would be great. thankyouuuuuuuu i hate this whole Ex.
p.s. im probably gonna reply with a couple more questions because there isnt a lot of explanations really, or i dont get them. but thankyou

[kiwiberry]

hello again i have a lot of questions about section 3.2 of 4u cambridge, mostly the parts involving multiple parametric points on ellipses or hyperbola. so the parts with chords between points: P (acosß, bsinß) and Q (acos∂, bsin∂). firstly, in the explanation/ working to show the lines equation and gradient there are a lot of steps missing and i cant understand how they get anywhere, mostly the gradients? secondly, how relevant/do we need to know this at all because my teacher doesnt even know how to do it really i dont think? and if we need to know, is it at a level of deriving, im gonna assume so because it cant hurt but still?

so yea if anyone wants to shoot up some kind of working for the gradients that would be great. thankyouuuuuuuu i hate this whole Ex.
p.s. im probably gonna reply with a couple more questions because there isnt a lot of explanations really, or i dont get them. but thankyou

They've used the sums to products formulas to simplify the gradients

[beau77bro]

Hey could please I get some help on these

1) P(cp,c/p) and Q(-cp,-c/p) are points on the rectangular hyperbola xy=c². The normal at P to the hyperbola meets the curve again at R. Find the coordinates of R, and hence show that \(\angle PQR\) is a right angle

2) P(3p,3/p) and Q(3q,3/q) are points on different branches of the hyperbola xy=9. The tangents at P and Q meet at T. Find the locus of T if the positions of P and Q vary so that the chord PQ passes through (0,4).
I managed to find the locus as x=9/2, but the answer says that it is x=9/2, y<0. I was wondering why y has to be less than 0? T is \(( \frac{6pq}{p+q} ,\frac{6}{p+q})\) for reference

Thank you!

ok so see the photos for q2:

but pretty much y<0 because the point p and q are on separate branches of a standard xy = c^2, so if we say q is on the branch in the 3rd quadrant, then its coordinates are all negative Q (3q, 3/q) and hence q is negative.
 
and when u rearrange  3(p+q) = 4pq
into p + q = 4pq/3

and sub into y = 6/(p + q)
you get:  y = 9/2pq   

and since p is positive and q is negative, y cannot be greater than 0, and it cant equal 0 because there isnt a variable on the top of the fraction.

[beau77bro]

They've used the sums to products formulas to simplify the gradients
(Image removed from quote.)

uhhhhh i dont think ive learnt this, how do you get these equations? do we need to know how to derive?

[kiwiberry]

ok so see the photos for q2:

but pretty much y<0 because the point p and q are on separate branches of a standard xy = c^2, so if we say q is on the branch in the 3rd quadrant, then its coordinates are all negative Q (3q, 3/q) and hence q is negative.
 
and when u rearrange  3(p+q) = 4pq
into p + q = 4pq/3

and sub into y = 6/(p + q)
you get:  y = 9/2pq   

and since p is positive and q is negative, y cannot be greater than 0, and it cant equal 0 because there isnt a variable on the top of the fraction.

omg that makes sense, thanks so much!!

[kiwiberry]

uhhhhh i dont think ive learnt this, how do you get these equations? do we need to know how to derive?

I'm pretty sure you don't have to derive them to use them, but there's a proof in 3U year12 cambridge I believe

[RuiAce]

uhhhhh i dont think ive learnt this, how do you get these equations? do we need to know how to derive?

[QC]

Hi, can someone help me with this question?
I feel like I know how to do it and I just suck at algebra or something. I've tried both realising the denominator and just forming one fraction, cross multiplying then solving the complex and real components separately as real and imaginary components.
Thanks

[kiwiberry]

Hi, can someone help me with this question?
I feel like I know how to do it and I just suck at algebra or something. I've tried both realising the denominator and just forming one fraction, cross multiplying then solving the complex and real components separately as real and imaginary components.
Thanks

you can equate real and imaginary parts from here