Author Topic: 3U Maths Question Thread (Read 1238024 times) Tweet Share

clovvy · « **Reply #3720 on:** October 14, 2018, 10:44:58 pm »

Quote from: RuiAce on October 14, 2018, 10:39:17 pm

$BCDE\text{ is a cyclic quadrilateral because the}\\ \text{interval }BC\text{ subtends the angles }\angle BDC\text{ and }\angle BEC,\\ \text{which are equal (since they are both altitudes and hence are both }90^\circ).$
$\text{Now draw the circle around }B,\,C,\,D\text{ and }E.$
$\text{Since }\angle BDC = \angle BEC = 90^\circ\text{, they are angles in a semicircle.}\\ \text{Therefore }CB\text{ is the diameter of the circle.}\\ \text{But then }M\text{ is the midpoint of a diameter.}\\ \text{Thus }M\text{ is in fact the circle's centre.}$
$\text{Now construct the lines }MD\text{ and }ME\text{, which are equal radii.}\\ \text{Consider }\triangle MNE\text{ and }\triangle MND.$
$\begin{align*}MD &= ME &\tag{from above}\\ MN &= MN \tag{common side}\\ NE &= ND\tag{N is the midpoint of DE} \end{align*}\\ \text{Hence }\triangle MND \equiv \triangle MNE\text{ by }SSS.$
$\text{Therefore }\angle MND = \angle MNE\text{ as they are matching}\\ \text{angles on congruent triangles.}\\ \text{But we also know that }\angle MNE = 180^\circ.$
$\text{Thus }\angle MND = \angle MNE = 90^\circ\text{, proving the desired result.}$

You're too quick -_-... I was typing stuff out to edit my reply T_T

iktimal · « **Reply #3721 on:** October 14, 2018, 11:04:48 pm »

Quote from: RuiAce on October 14, 2018, 10:39:17 pm

$BCDE\text{ is a cyclic quadrilateral because the}\\ \text{interval }BC\text{ subtends the angles }\angle BDC\text{ and }\angle BEC,\\ \text{which are equal (since they are both altitudes and hence are both }90^\circ).$
$\text{Now draw the circle around }B,\,C,\,D\text{ and }E.$
$\text{Since }\angle BDC = \angle BEC = 90^\circ\text{, they are angles in a semicircle.}\\ \text{Therefore }CB\text{ is the diameter of the circle.}\\ \text{But then }M\text{ is the midpoint of a diameter.}\\ \text{Thus }M\text{ is in fact the circle's centre.}$
$\text{Now construct the lines }MD\text{ and }ME\text{, which are equal radii.}\\ \text{Consider }\triangle MNE\text{ and }\triangle MND.$
$\begin{align*}MD &= ME &\tag{from above}\\ MN &= MN \tag{common side}\\ NE &= ND\tag{N is the midpoint of DE} \end{align*}\\ \text{Hence }\triangle MND \equiv \triangle MNE\text{ by }SSS.$
$\text{Therefore }\angle MND = \angle MNE\text{ as they are matching}\\ \text{angles on congruent triangles.}\\ \text{But we also know that }\angle MNE = 180^\circ.$
$\text{Thus }\angle MND = \angle MNE = 90^\circ\text{, proving the desired result.}$

I don't understand why " Since < BDC = < BEC = 90, they are angles in a semicircle", can you please explain to me why they are angles in a semicircle not a full circle?

RuiAce · « **Reply #3722 on:** October 14, 2018, 11:07:04 pm »

Quote from: iktimal on October 14, 2018, 11:04:48 pm

I don't understand why " Since < BDC = < BEC = 90, they are angles in a semicircle", can you please explain to me why they are angles in a semicircle not a full circle?

I am trying to use the converse of the theorem "the angle in a semicircle is a right angle". The whole point of that step was to prove that $BC$ is a diameter, which I cannot do without using (the converse of) this theorem.

Because the theorem says the word "semicircle" in it, I used the word "semicircle" as well.

iktimal · « **Reply #3723 on:** October 14, 2018, 11:12:06 pm »

Quote from: RuiAce on October 14, 2018, 11:07:04 pm

I am trying to use the converse of the theorem "the angle in a semicircle is a right angle". The whole point of that step was to prove that $BC$ is a diameter, which I cannot do without using (the converse of) this theorem.

Because the theorem says the word "semicircle" in it, I used the word "semicircle" as well.

Oh OK, Thank you very much

siimraan · « **Reply #3724 on:** October 16, 2018, 11:23:20 pm »

Quote from: RuiAce on October 14, 2018, 10:39:17 pm

$BCDE\text{ is a cyclic quadrilateral because the}\\ \text{interval }BC\text{ subtends the angles }\angle BDC\text{ and }\angle BEC,\\ \text{which are equal (since they are both altitudes and hence are both }90^\circ).$
$\text{Now draw the circle around }B,\,C,\,D\text{ and }E.$
$\text{Since }\angle BDC = \angle BEC = 90^\circ\text{, they are angles in a semicircle.}\\ \text{Therefore }CB\text{ is the diameter of the circle.}\\ \text{But then }M\text{ is the midpoint of a diameter.}\\ \text{Thus }M\text{ is in fact the circle's centre.}$
$\text{Now construct the lines }MD\text{ and }ME\text{, which are equal radii.}\\ \text{Consider }\triangle MNE\text{ and }\triangle MND.$
$\begin{align*}MD &= ME &\tag{from above}\\ MN &= MN \tag{common side}\\ NE &= ND\tag{N is the midpoint of DE} \end{align*}\\ \text{Hence }\triangle MND \equiv \triangle MNE\text{ by }SSS.$
$\text{Therefore }\angle MND = \angle MNE\text{ as they are matching}\\ \text{angles on congruent triangles.}\\ \text{But we also know that }\angle MNE = 180^\circ.$
$\text{Thus }\angle MND = \angle MNE = 90^\circ\text{, proving the desired result.}$

After you prove BC is the diameter and M is the centre. Since DE is a chord and N is the midpoint cant you just use the theorem that 'The line from the centre of a circle to the midpoint of a chord is perpendicular to the chord.' Or do you have to do the extra working to get the full marks.

RuiAce · « **Reply #3725 on:** October 16, 2018, 11:36:31 pm »

Quote from: siimraan on October 16, 2018, 11:23:20 pm

After you prove BC is the diameter and M is the centre. Since DE is a chord and N is the midpoint cant you just use the theorem that 'The line from the centre of a circle to the midpoint of a chord is perpendicular to the chord.' Or do you have to do the extra working to get the full marks.

Fair call. I forgot that theorem existed because I've never seen it get used until now.

rachtrenaman · « **Reply #3726 on:** October 22, 2018, 06:34:01 pm »

Can someone please help with this question. I’m seriously struggling with locus and parabola questions!

The chord of contact of the tangents to the parabola x²= 4ay from an external point P(x₀,y₀) passes through point Q(0,2a). Find the equation of the locus of the midpoint of PQ.
Thanks

RuiAce · « **Reply #3727 on:** October 22, 2018, 07:05:56 pm »

Quote from: rachtrenaman on October 22, 2018, 06:34:01 pm

Can someone please help with this question. I’m seriously struggling with locus and parabola questions!

The chord of contact of the tangents to the parabola x²= 4ay from an external point P(x₀,y₀) passes through point Q(0,2a). Find the equation of the locus of the midpoint of PQ.
Thanks

$\text{Recall from the formula sheet that the chord of contact}\\ \text{has the equation }xx_0 = 2a(y+y_0).\\ \text{If it passes through }(0,2a)\text{, subbing this in gives}\\ 0 = 2a(2a + y_0) \implies \boxed{y_0 = -2a}.$
$\text{Now the midpoint of }PQ\text{ is the point }\boxed{\left( \frac{x_0}{2}, \frac{y_0+2a}{2} \right)}\\ \text{and thus satisfies the two conditions}\\ \begin{align*}x&=\frac{x_0}{2}\\ y&= \frac{y_0+2a}{2} \end{align*}$
$\text{Since }y_0 = -2a\text{, the second equation becomes }\boxed{y=0}.\\ \text{The parameter values }x_0\text{ and }y_0\text{ are already eliminated in this equation}\\ \text{and hence the locus of the midpoint is just }\boxed{y = 0}\text{, i.e. the }x\text{-axis.}$

rachtrenaman · « **Reply #3728 on:** October 22, 2018, 08:13:03 pm »

Quote from: RuiAce on October 22, 2018, 07:05:56 pm

$\text{Recall from the formula sheet that the chord of contact}\\ \text{has the equation }xx_0 = 2a(y+y_0).\\ \text{If it passes through }(0,2a)\text{, subbing this in gives}\\ 0 = 2a(2a + y_0) \implies \boxed{y_0 = -2a}.$
$\text{Now the midpoint of }PQ\text{ is the point }\boxed{\left( \frac{x_0}{2}, \frac{y_0+2a}{2} \right)}\\ \text{and thus satisfies the two conditions}\\ \begin{align*}x&=\frac{x_0}{2}\\ y&= \frac{y_0+2a}{2} \end{align*}$
$\text{Since }y_0 = -2a\text{, the second equation becomes }\boxed{y=0}.\\ \text{The parameter values }x_0\text{ and }y_0\text{ are already eliminated in this equation}\\ \text{and hence the locus of the midpoint is just }\boxed{y = 0}\text{, i.e. the }x\text{-axis.}$

thank you!

chayek · « **Reply #3729 on:** October 22, 2018, 08:23:55 pm »

Hi, I am having some issues trying to solve this parametrics question. I don't understand what to do and where to start, if someone could please help.

"If the chord of contact to the tangents of the parabola x^2 = 4ay, from an external point T(x1,y1) meets the directrix at R, prove that RT subtends a right angle at the focus

thanks

RuiAce · « **Reply #3730 on:** October 22, 2018, 08:38:09 pm »

Quote from: chayek on October 22, 2018, 08:23:55 pm

Hi, I am having some issues trying to solve this parametrics question. I don't understand what to do and where to start, if someone could please help.

"If the chord of contact to the tangents of the parabola x^2 = 4ay, from an external point T(x1,y1) meets the directrix at R, prove that RT subtends a right angle at the focus

thanks

The diagram gives you all important pieces of the puzzle. (Note that the tangents aren't really useful and hence dimmde out; they just reinforce the fact that the other line is indeed the chord of contact.

You are trying to prove that $\angle TSR = 90^\circ$.

My first instinct (may or may not work) for this question would be to use the formula sheet and the fact that perpendicular lines have gradient $m_1m_2 = -1$.

Divayth Fyr · « **Reply #3731 on:** October 23, 2018, 04:51:29 pm »

Quote from: chayek on October 22, 2018, 08:23:55 pm

Hi, I am having some issues trying to solve this parametrics question. I don't understand what to do and where to start, if someone could please help.

"If the chord of contact to the tangents of the parabola x^2 = 4ay, from an external point T(x1,y1) meets the directrix at R, prove that RT subtends a right angle at the focus

thanks

$\text{From the reference sheet, the equation for the chord of contact is: } xx_1 = 2a(y+y_1) \\ \text{The point } R \ \text{lies on the directrix, and so it has y coordinate } = -a \\ \text{Substituting this into the equation for the chord of contact, we get: } \\ \begin{align*} xx_1 &= 2a(-a + y_1) \\ x &= \frac{2a(y_1-a)}{x_1} \\ \end{align*} \\ \therefore R\ \text{has cordinates } \left(\frac{2a(y_1-a)}{x_1}, -a \right) \\ \text{Let the focus of the parabola be } S. \\ S\ \text{has cordinates } (0, a) \\ \text{We want to show that } TR\ \text{makes a right angle at } S. \\ \text{We have coordiantes for all these points, so the easiest way is to show that } TS\ \text{and } RS\ \text{are perpendicular i.e. } m_{TS}\times m_{RS} = -1 $
$\begin{align*} m_{TS} &= \frac{y_1-a}{x_1-0} \\ &= \frac{y_1-a}{x_1} \\ \end{align*}$
$\begin{align*} m_{RS} &= \frac{-a-a}{\frac{2a(y_1-a)}{x_1}} \\ &= \frac{-2ax_1}{2a(y_1-a)} \\ &= \frac{-x_1}{y_1-a} \\ \end{align*}$
$\begin{align*} \therefore m_{TS}\times m_{RS} &= \frac{-x_1}{y_1-a} \times \frac{y_1-a}{x_1} \\ &= -1 \end{align*} \\ \therefore RT\ \text{subtends a right angle at the focus.}$

Mate2425 · « **Reply #3732 on:** October 26, 2018, 08:21:56 am »

Hi, i was wondering how you prove that a point given, is a point of inflexion, if it lies on a curve?

Thanks

RuiAce · « **Reply #3733 on:** October 26, 2018, 08:29:22 am »

Quote from: Mate2425 on October 26, 2018, 08:21:56 am

Hi, i was wondering how you prove that a point given, is a point of inflexion, if it lies on a curve?

Thanks

Show that at that point the second derivative is 0, and then test both sides to ensure there’s a change in concavity.

Mate2425 · « **Reply #3734 on:** October 26, 2018, 11:49:02 am »

Hey also with asin@ + bcos@ = rsin(@+&) could you please tell me what the special rule is where if they change position to this standard (<--) or one term has a negative in front of it instead of a +. What sort of process should i take to tackle those sort of questions?

Thaaaannnnkkkyyyooouu!

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