 February 25, 2020, 04:16:23 pm

### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1194393 times) Tweet Share

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#### TheEagle

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« Reply #9600 on: January 25, 2020, 11:31:18 am »
+1
Edit: Sorry I realized I made a mistake, tbh I am also not sure why it is positive a, so let's just wait till help comes.

no worries
thanks for offering

#### cotangent ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9601 on: January 25, 2020, 02:47:11 pm »
+1
Hello all

I am having trouble finding the range of x over the domain of (pi/2, 3pi/2). I keep getting negative infinity to negative a, but the textbook says negative infinity to positive a?

I think they made an error, it should be -infinity to -a.
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#### ^^^111^^^

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« Reply #9602 on: January 25, 2020, 03:09:10 pm »
+1
I think they made an error, it should be -infinity to -a.
Same I believe they made a mistake as well.

#### TheEagle

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« Reply #9603 on: January 25, 2020, 06:16:49 pm »
+1
Seems like it. When I started doing the exercises, there were questions that asked for the domain and followed the same method (-negative infinity to a)

#### MaiSakurajima ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9604 on: January 29, 2020, 02:57:30 am »
0
Hey guys can I get some help on this question Extended response:
1. e. Find the equation of the locus of M when:
i. b =/  1/sqrt(2)
ii b = 1/sqrt(2)

What is this question asking for? Like I am currently stumped.
Thanks
https://imgur.com/a/3VfKonZ
https://imgur.com/a/0z04QjS

uh its not letting me attach images i am sorry
« Last Edit: January 29, 2020, 03:02:01 am by MaiSakurajima »
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#### Bri MT

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« Reply #9605 on: January 29, 2020, 07:44:45 am »
0
Hey guys can I get some help on this question Extended response:
1. e. Find the equation of the locus of M when:
i. b =/  1/sqrt(2)
ii b = 1/sqrt(2)

What is this question asking for? Like I am currently stumped.
Thanks
https://imgur.com/a/3VfKonZ
https://imgur.com/a/0z04QjS

uh its not letting me attach images i am sorry

The reason it's not letting you attach the image is because you aren't giving it the right url. Got to the top right corner of the image and lick to open the menu. Then select "get share links" and copy the bbcode over to the forums.

You can test if you have the right link by clicking on the url, if it doesn't show you only that image the url is wrong.

Don't need to apologise for that though, embedding images can be confusing 2018-2021: Science Advanced - Global Challenges (Honours) @ Monash

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Want QCE help? Leave a post here #### dream chaser

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« Reply #9606 on: February 03, 2020, 06:39:00 pm »
0
Hi Guys,

I need help with this question. How would you do it?

Find the equation of the tangent at the point (2, 1) of the circle x^2 + y^2 − 4y − 1 = 0

I know the equation is y=m(x-2)+1 from simply subbing (2,1) into y-y1=m(x-x1) formula. I just need to find m now but can't fully solve it.

I subbed y=m(x-2)+1 into x^2+y^2-4y-1=0 and ended up with: x^2 +(mx-2m+1)^2 - 4(mx-2m+1) -1=0. What do I do now?

All help will be much appreciated. Thanks

PS: Sorry, I just realized I accidentally post the message twice from looking at the forum. My apologies for that. I thought it didn't get sent the first time around.

#### jnlfs2010 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9607 on: February 03, 2020, 07:17:33 pm »
+1
Hi Guys,

I need help with this question. How would you do it?

Find the equation of the tangent at the point (2, 1) of the circle x^2 + y^2 − 4y − 1 = 0

I know the equation is y=m(x-2)+1 from simply subbing (2,1) into y-y1=m(x-x1) formula. I just need to find m now but can't fully solve it.

I subbed y=m(x-2)+1 into x^2+y^2-4y-1=0 and ended up with: x^2 +(mx-2m+1)^2 - 4(mx-2m+1) -1=0. What do I do now?

All help will be much appreciated. Thanks

PS: Sorry, I just realized I accidentally post the message twice from looking at the forum. My apologies for that. I thought it didn't get sent the first time around.

You've got a quadratic. Now get into ax2+bx+c=0 form. Since the line is a tangent, the discriminant of that equation must equal to 0 as there is only one solution. With the value of m, you surely should get the formula of the tangent
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#### dream chaser

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« Reply #9608 on: February 03, 2020, 07:24:11 pm »
0
You've got a quadratic. Now get into ax2+bx+c=0 form. Since the line is a tangent, the discriminant of that equation must equal to 0 as there is only one solution. With the value of m, you surely should get the formula of the tangent

But what about the x^2 in front? What am I meant to do with that? Thanks for the reply by the way.

#### ^^^111^^^

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« Reply #9609 on: February 05, 2020, 12:45:22 pm »
0
But what about the x^2 in front? What am I meant to do with that? Thanks for the reply by the way.
You can always use calculus. dy/dx of x2+ y2-4y-1 is -x/y-2
Then we can use our coordinates to (2,1) to get dy/dx or gradient is equal to 2. Then find the y intercept of (1)=2(2)+c which gives c=-3. Thus equation of tangent is y=2x-3

#### dream chaser

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« Reply #9610 on: February 05, 2020, 12:58:03 pm »
0
You can always use calculus. dy/dx of x2+ y2-4y-1 is -x/y-2
Then we can use our coordinates to (2,1) to get dy/dx or gradient is equal to 2. Then find the y intercept of (1)=2(2)+c which gives c=-3. Thus equation of tangent is y=2x-3

Thanks for the reply. This is the solution the book gave. Could someone explain it to me please how the book got it.

For instance, what does it mean about 2 equal roots and why is that the case here?

#### fun_jirachi

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« Reply #9611 on: February 06, 2020, 11:29:44 am »
+3
Hey there!

When jnlfs2010 said this:
You've got a quadratic. Now get into ax2+bx+c=0 form. Since the line is a tangent, the discriminant of that equation must equal to 0 as there is only one solution. With the value of m, you surely should get the formula of the tangent

They essentially implied expanding and simplifying the quadratic you ended up with, which was $x^2 +(mx-2m+1)^2 - 4(mx-2m+1) -1=0$, which also happened to be book's solution. They also told you to use the discriminant, and let it equal to zero.

Essentially, when a line cuts the parabola, you have a maximum of two solutions, and a minimum of zero. Equating the line and the parabola will give you the x-values of the points of intersection as roots of a quadratic. If the quadratic is a perfect square, then $\Delta = 0$, and there is one real root ie. the line is a tangent to the parabola, and does not in fact cut the parabola in two different places like when the discriminant is greater than zero ie. the 'two roots' are equal - there is thus only one real root!

Since we want to find the tangent to the circle at (2, 1), there is only one solution (or if you want to think about it like that, there are two equal solutions!) since there is only one point of contact. Hence, the book (and we!) choose to expand the brackets in this quadratic in x $x^2 +(mx-2m+1)^2 - 4(mx-2m+1) -1=0$ , then find the discriminant using $\Delta = b^2 - 4ac$, and let it equal zero, since we only want the one solution. They've got in the book that b2 = 4ac, which is the same as b2 - 4ac equals zero. Since there is only one solution, we expect a perfect square with a quadratic in m, thus finding the gradient of the tangent, then using that to find the equation of the line using point gradient form.

A quicker alternative method if you've learned it is suggested by ^^^111^^^, implicit differentiation. Implicit differentiation is just an extension of the chain rule, ie.

In our case, we have that

Isolating $\frac{dy}{dx}$ and substituting the coordinates of any point on the given circle will give the gradient of the tangent at that point, just like differentiating any other curve. From here, the steps are the same as the method given in the book.

Hope this clears everything up! Failing everything, but I'm still Flareon up.

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