Author Topic: 4U Maths Question Thread (Read 665414 times) Tweet Share

Mahan · « **Reply #1110 on:** April 24, 2017, 12:48:36 am »

Quote from: beau77bro on April 23, 2017, 12:13:29 am

(Image removed from quote.)

Please help me question 4c, I feel I've done it before, but a solution would be so helpful

Using the first part,
$A=\frac{a^{2}-b^{2}}{a}cos\theta, B=\frac{b^{2}-a^{2}}{b}sin\theta$
Also as a reminder, $\frac{b^{2}}{a^{2}} = 1-e^{2}$
Note
$PA^{2} = (a \ cos\theta - \frac{a^{2}-b^{2}}{a}cos\theta) ^{2} + b^{2} \ sin^{2}\theta = b^{2} \ sin^{2}\theta+\frac{b^{4}}{a^{2}}cos^{2}\theta$
Similarly
$PB^{2} = a^{2} \ cos^{2}\theta + (b \ sin\theta - \frac{b^{2}-a^{2}}{b}sin\theta )^{2} = a^{2} \ cos^{2}\theta + \frac{a^{4}}{b^{2}} sin^{2}\theta$
Then
$\frac{PA}{PB} = \frac{b}{a} \sqrt{1-e^{2}} = 1-e^{2}$
This comes from the facts:
$\frac{b^{2}}{a^{2}} = 1-e^{2}$
and
$PB^{2} = a^{2} \ cos^{2}\theta + \frac{a^{4}}{b^{2}} sin^{2}\theta = \frac{a^{2}b^{2} cos^{2}\theta + a^{4}sin^{2}\theta}{b^{2}} = \frac{a^{4}}{b^{2}} ((1-e^{2})cos^{2}\theta + sin^{2}\theta)$
and
$a^{4} \ (1-e^{2}) \frac{sin^{2} \theta + (1-e^{2}) cos^{2}\theta}{a^{2}} = PA^{2}$

hanaacdr · « **Reply #1111 on:** April 24, 2017, 12:40:58 pm »

Hi
could i get some help on this question please
i feel like I have done it before but not getting the answer
thank you!

hanaacdr · « **Reply #1112 on:** April 24, 2017, 12:43:34 pm »

Hi
Sorry and this question please
I feel like it is very physicsy (hhaha not a word)
thank you!

hanaacdr · « **Reply #1113 on:** April 25, 2017, 12:15:35 pm »

hi
Could i please get some help on question 8 please
thank youuu

jakesilove · « **Reply #1114 on:** April 25, 2017, 12:42:37 pm »

Quote from: hanaacdr on April 24, 2017, 12:40:58 pm

Hi
could i get some help on this question please
i feel like I have done it before but not getting the answer
thank you!

Hey! For this question, simply use the distance formula to find the distance between the points

$P(a\cos\theta, b\sin\theta), (ae, 0), (-ae,0)$

Where the eccentricity takes its usual form. You'll find that the distance between the point and the 'positive' foci is

$a(1-e\cos\theta)$

and the distance between the point and the 'negative' foci is

$a(1+e\cos\theta)$

Adding this up gives us the required relation, 2a.

Quote from: hanaacdr on April 24, 2017, 12:43:34 pm

Hi
Sorry and this question please
I feel like it is very physicsy (hhaha not a word)
thank you!

I'm not sure that this is really a maths question; I think it wants you to say that the beam of light will again go through the first foci? Presumably, EVERY bounce will lead the light to one of the foci. Nothing to solve here

RuiAce · « **Reply #1115 on:** April 25, 2017, 12:51:50 pm »

Quote from: hanaacdr on April 24, 2017, 12:43:34 pm

Hi
Sorry and this question please
I feel like it is very physicsy (hhaha not a word)
thank you!

The ray is going to alternate through which focus it passes through. If you shine a light from, or directly through one focus, it will always be reflected to the other focus. This is the reflection property of the ellipse.

But yeah, it's probably more of a physics question.

Quote from: jakesilove on April 25, 2017, 12:42:37 pm

Hey! For this question, simply use the distance formula to find the distance between the points

$P(a\cos\theta, b\sin\theta), (ae, 0), (-ae,0)$

Where the eccentricity takes its usual form. You'll find that the distance between the point and the 'positive' foci is

$a(1-e\cos\theta)$

and the distance between the point and the 'negative' foci is

$a(1+e\cos\theta)$

It says Cartesian coordinates in the question. Give me a sec.

RuiAce · « **Reply #1116 on:** April 25, 2017, 12:56:02 pm »

Quote from: hanaacdr on April 24, 2017, 12:40:58 pm

Hi
could i get some help on this question please
i feel like I have done it before but not getting the answer
thank you!

$\text{The standard method is just this}$
$\begin{align*}PS+PS^\prime &=ePM + ePM^\prime \tag{focus-directrix definition}\\ &= e\left(PM+PM^\prime\right)\\ &= eMM^\prime\\&=e \left(\frac{2a}{e}\right)\\ &= 2a\end{align*}$
$\text{The trick is that }PM\text{ and }PM^\prime\text{ are just collinear horizontal lines}\\ \text{This should be clear if you draw a diagram}$

RuiAce · « **Reply #1117 on:** April 25, 2017, 01:01:37 pm »

Quote from: hanaacdr on April 25, 2017, 12:15:35 pm

hi
Could i please get some help on question 8 please
thank youuu

I just simulated it on GeoGebra and the question is wrong. (Most likely a typo.)

Because ellipse ii is literally enclosed INSIDE ellipse i, a tangent to ellipse i can be shown to never touch ellipse ii.

chelseam · « **Reply #1118 on:** April 25, 2017, 10:15:08 pm »

Hi! I'm confused about how to prove the reflection property for the hyperbola! Also, would the geometrical properties for the hyperbola also apply to rectangular hyperbolae (especially the type with its asymptotes being the x & y axis)? Thank you

RuiAce · « **Reply #1119 on:** April 25, 2017, 11:16:26 pm »

Quote from: chelseam on April 25, 2017, 10:15:08 pm

Hi! I'm confused about how to prove the reflection property for the hyperbola! Also, would the geometrical properties for the hyperbola also apply to rectangular hyperbolae (especially the type with its asymptotes being the x & y axis)? Thank you

A rectangular hyperbola is just a hyperbola with eccentricity √2, so any property with the hyperbola will always apply to the rectangular hyperbola.

That being said, the foci and directrices of $xy=c^2$ are NOT required for the 4U course, so you should NOT have to expect to do much with them.

For a full proof of the reflection property, consider Q13b and Q13c of the 2012 paper as well as any good source of solutions. You may ask for any specific lines of working out to be explained.

hanaacdr · « **Reply #1120 on:** April 26, 2017, 10:16:28 pm »

Quote from: RuiAce on April 25, 2017, 01:01:37 pm

I just simulated it on GeoGebra and the question is wrong. (Most likely a typo.)

Because ellipse ii is literally enclosed INSIDE ellipse i, a tangent to ellipse i can be shown to never touch ellipse ii.

Thank you!

JuliaPascale123 · « **Reply #1121 on:** April 27, 2017, 11:29:02 am »

Could I please have the working out and answers to these questions

RuiAce · « **Reply #1122 on:** April 27, 2017, 12:26:59 pm »

Quote from: JuliaPascale123 on April 27, 2017, 11:29:02 am

Could I please have the working out and answers to these questions
(Image removed from quote.)

$\text{Observing the need for the product rule for the second term}\\ \text{By implicit differentiation}$
$\begin{align*}x^2-xy+y^2&=3\\ 2x - \left(x\frac{dy}{dx}+y\right) + 2y\frac{dy}{dx}&=0\\ (2y-x)\frac{dy}{dx}&= y-2x\\ \frac{dy}{dx}&=\frac{y-2x}{2y-x}\\ &= \frac{2x-y}{x-2y}\end{align*}$

beau77bro · « **Reply #1123 on:** April 27, 2017, 07:44:18 pm »

Quote from: Mahan on April 24, 2017, 12:48:36 am

Using the first part,
$A=\frac{a^{2}-b^{2}}{a}cos\theta, B=\frac{b^{2}-a^{2}}{b}sin\theta$
Also as a reminder, $\frac{b^{2}}{a^{2}} = 1-e^{2}$
Note
$PA^{2} = (a \ cos\theta - \frac{a^{2}-b^{2}}{a}cos\theta) ^{2} + b^{2} \ sin^{2}\theta = b^{2} \ sin^{2}\theta+\frac{b^{4}}{a^{2}}cos^{2}\theta$
Similarly
$PB^{2} = a^{2} \ cos^{2}\theta + (b \ sin\theta - \frac{b^{2}-a^{2}}{b}sin\theta )^{2} = a^{2} \ cos^{2}\theta + \frac{a^{4}}{b^{2}} sin^{2}\theta$
Then
$\frac{PA}{PB} = \frac{b}{a} \sqrt{1-e^{2}} = 1-e^{2}$
This comes from the facts:
$\frac{b^{2}}{a^{2}} = 1-e^{2}$
and
$PB^{2} = a^{2} \ cos^{2}\theta + \frac{a^{4}}{b^{2}} sin^{2}\theta = \frac{a^{2}b^{2} cos^{2}\theta + a^{4}sin^{2}\theta}{b^{2}} = \frac{a^{4}}{b^{2}} ((1-e^{2})cos^{2}\theta + sin^{2}\theta)$
and
$a^{4} \ (1-e^{2}) \frac{sin^{2} \theta + (1-e^{2}) cos^{2}\theta}{a^{2}} = PA^{2}$

omg thankyou soo much mahan, sorry such a late reply appreciate the make explanation

beau77bro · « **Reply #1124 on:** April 27, 2017, 07:46:08 pm »

This is an absolute monster question. I think atleast ahahaha any help is ridiculously appreciated

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