Author Topic: 4U Maths Question Thread (Read 665079 times) Tweet Share

RuiAce · « **Reply #810 on:** February 07, 2017, 09:32:27 pm »

Quote from: aroon217 on February 07, 2017, 09:26:48 pm

oh dang I didn't apply de Moivres again with the (costheta+isintheta)^2 only the power n so i was getting random half powers and I couldn't get it. Thanks man, its pretty annoying this stuff isnt online

You said this was a past HSC paper question?

hinakamishiro · « **Reply #811 on:** February 10, 2017, 05:36:20 pm »

Hey guys could i have some help with part b please? Thanks

RuiAce · « **Reply #812 on:** February 10, 2017, 06:03:09 pm »

Quote from: hinakamishiro on February 10, 2017, 05:36:20 pm

Hey guys could i have some help with part b please? Thanks

$\text{Note that }\boxed{\tan 4\theta=\frac{\sin 4\theta}{\cos 4\theta}}$
$\text{To find }\sin 4\theta\text{ and }\cos 4\theta\text{ without double angles}\\ \text{Note that by combining De Moivre's theorem with the binomial theorem}\\ \begin{align*}\cos 4\theta+i\sin 4\theta&=(\cos \theta+i\sin \theta)^4\\ &=\cos^4\theta+4i\cos^3\theta\sin\theta-6\cos^2\theta\sin^2\theta-4i\cos\theta\sin^3\theta+\sin^4\theta\end{align*}$
$\text{Hence by equating the real and imaginary parts}\\ \begin{align*}\cos 4\theta&=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\\ \sin 4\theta&=4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta\end{align*}$
$\text{Subbing this into the above expression and then dividing by }\cos^4\theta\text{ top and bottom}\\ \begin{align*}\tan 4\theta&=\frac{4\cos^3\theta\sin\theta-4\cos\theta\sin^3}{\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta}\\ &= \frac{4\tan\theta-4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta}\end{align*}$

Rathin · « **Reply #813 on:** February 12, 2017, 03:22:34 pm »

Find the equation of the tangents to the ellipse 4x^2+9y^2=36 given that these tangents are perpendicular to the line 3x+2y=5.

I got 2x-3y+6=0
Answer is 2x-3y+6sqrt(2)=0

RuiAce · « **Reply #814 on:** February 12, 2017, 03:25:53 pm »

Quote from: Rathin on February 12, 2017, 03:22:34 pm

Find the equation of the tangents to the ellipse 4x^2+9y^2=36 given that these tangents are perpendicular to the line 3x+2y=5.

I got 2x-3y+6=0
Answer is 2x-3y+6sqrt(2)=0

Judging by what the question asks it's likely you made a little mistake somewhere in your working, especially since you were only off by a factor. Can you please post it up?

Rathin · « **Reply #815 on:** February 12, 2017, 03:39:20 pm »

Quote from: RuiAce on February 12, 2017, 03:25:53 pm

Judging by what the question asks it's likely you made a little mistake somewhere in your working, especially since you were only off by a factor. Can you please post it up?

I have done it completely wrong.

RuiAce · « **Reply #816 on:** February 12, 2017, 03:40:00 pm »

Rathin · « **Reply #817 on:** February 12, 2017, 03:57:29 pm »

Quote from: RuiAce on February 12, 2017, 03:40:00 pm

$\text{Actually, hold that suggestion. I just thought of something more interesting.}$
$\text{Any line perpendicular to }3x+2y=5\text{ will take the form}\\ 2x-3y+c=0\\ \text{Where }c\text{ is a constant in the equation }ax+by+c=0$
$\text{Rearranging this equation yields}\\ y=\frac23x+\frac{c}3$
$\text{Suppose we want to find the point of intersection}\\ \text{Between this line and the ellipse}\\ \text{By substituting:}\\ \begin{align*}4x^2+9\left(\frac23x+\frac{c}3\right)^2&=36\\ 4x^2+(2x+c)^2&=36\\ 8x^2+(4c)x+(c^2-36)&=0\end{align*}$
$\text{This quadratic equation should only have one solution for }x\\ \text{That is, if the line is TANGENT to the ellipse}\\ \textbf{Note that a tangent to an ellipse only ever intersects it once}\\ \text{which is at the point where the line IS tangential}$
$\text{If the equation only has one solution AND is a quadratic}\\ \text{The discriminant should be }0\\ \begin{align*}16c^2-32(c^2-36)&=0\\ c^2&=72\\ c&=\pm 6\sqrt2\end{align*}$
$\text{Hence our required lines are}\\ 2x+3y\pm 6\sqrt2=0$

Thanks, that was very smart

I was trying to equate the implicitly differentiated ellipse and the given gradient (2/3) to come out with a point what the gradient of 2/3 is true and then find the eqn of it but that didn't work out.

hanaacdr · « **Reply #818 on:** February 12, 2017, 05:17:56 pm »

Hi!!
could i get some help on this volumes question please,
i have done it but i keep getting the wrong answer
thank you!

RuiAce · « **Reply #819 on:** February 12, 2017, 05:42:36 pm »

Quote from: hanaacdr on February 12, 2017, 05:17:56 pm

Hi!!
could i get some help on this volumes question please,
i have done it but i keep getting the wrong answer
thank you!

Which method do you want? Shells is easier unless you haven't been taught how to use it.

hanaacdr · « **Reply #820 on:** February 12, 2017, 05:58:54 pm »

Yes i have been taught shells, but we were given this question to use the slicing method

RuiAce · « **Reply #821 on:** February 12, 2017, 06:38:54 pm »

Rathin · « **Reply #822 on:** February 17, 2017, 12:51:35 pm »

The ellipse x^2/a^2 + y^2/b^ =1 meets the y=axis at C and D. Tangents drawn to C and D on the ellipse meet the tangent in (xcos(θ))/a + (ysin(θ))/b=1 at the points E and F respectively. Prove that CE*DF=a^2

RuiAce · « **Reply #823 on:** February 17, 2017, 01:22:22 pm »

Quote from: Rathin on February 17, 2017, 12:51:35 pm

The ellipse x^2/a^2 + y^2/b^ =1 meets the y=axis at C and D. Tangents drawn to C and D on the ellipse meet the tangent in (xcos(θ))/a + (ysin(θ))/b=1 at the points E and F respectively. Prove that CE*DF=a^2

$\text{Clearly, }C\text{ is just the point }(0,b)\\ \text{and }D\text{ is the point }(0,-b)\\ \text{(Or other way around - it does not matter.)}$
$\text{The tangent at }C\text{ is clearly the horizontal line }y=b\\ \text{The tangent at }D\text{ is clearly the horizontal line }y=-b$
$\text{To kill 2 birds with 1 stone}\\ \text{Sub }y=\pm b\text{ into the tangent at }(a\cos \theta, b\sin \theta)\\ \text{to find both points of intersection:}$
$\begin{align*}\frac{x\cos \theta}a+\frac{y\sin \theta}b&=1\\ \frac{x\cos \theta}a + \frac{\pm b \sin \theta}b&=1\\ \frac{x\cos \theta}a \pm \sin \theta &=1\\ x&= \frac{a}{\cos \theta}\left(1\mp \sin \theta\right)\end{align*}$
$\therefore E\text{ is the point }\left(\frac{a}{\cos \theta}(1-\sin \theta), b\right)\\ \text{and }F\text{ is the point }\left(\frac{a}{\cos \theta}(1+\sin \theta,-b\right)$
$\text{Because of the fact }CE\text{ and }DF\text{ are horizontal distances}\\ CE = \left| \frac{a}{\cos \theta}(1-\sin \theta)-0\right|\\ DF = \left| \frac{a}{\cos \theta}(1+\sin \theta)-0\right|$
$\text{Hence}\\ \begin{align*}CE\times DF &=\left| \frac{a}{\cos \theta}(1-\sin \theta)\right| \left| \frac{a}{\cos \theta}(1+\sin \theta)\right|\\ &= a^2 \left|\frac{1-\sin^2\theta}{\cos^2\theta}\right|\\ &= a^2\end{align*}$

kinky_khan · « **Reply #824 on:** February 18, 2017, 06:14:29 pm »

HEYYYY Jake, I was struggling to answer the following questions and I was wondering if you could help me with them? thank you SOOOO MUCHH!!!

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