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Need help with questions please

4C82607D-7103-4C05-BCCA-F3A31EAD250F.jpeg
I have done the implicit differentiation required from part 6a) but I dont know where to start to complete part 6b)
Reply 1
There aren't really any shortcuts to finding distances - you need the coordinates of P and Q. That's the goal.
(Btw, it's a good idea to literally write down on your script "To find the distance, we wish to find the coordinates of P and Q". Probably this sentence worth no marks, but it's good for the marker, but more importantly for you.)

Let's start with finding P, as finding Q is practically the same. You know:
(i) the tangent to the curve at P is parallel to the y-axis. How does this relate to dy/dx (recall dy/dx is just the gradient of the tangent line)?
(ii) P lies on the curve. Using (i), can you set up an equation to find the coordinates of P?
(Again, it's a good idea to have some descriptive words like "since the tangent at P is parallel to the y-axis, we have 'blah blah blah and calculations'")
(edited 12 months ago)
Original post by Matheen1
4C82607D-7103-4C05-BCCA-F3A31EAD250F.jpeg
I have done the implicit differentiation required from part 6a) but I dont know where to start to complete part 6b)

Do you happen to have an answer as to what k is? If my long-winded method gives the correct answer, then I'll explain what I did, but I don't know if it's even right.
Reply 3
Original post by tonyiptony
There aren't really any shortcuts to finding distances - you need the coordinates of P and Q. That's the goal.
(Btw, it's a good idea to literally write down on your script "To find the distance, we wish to find the coordinates of P and Q". Probably this sentence worth no marks, but it's good for the marker, but more importantly for you.)

Let's start with finding P, as finding Q is practically the same. You know:
(i) the tangent to the curve at P is parallel to the y-axis. How does this relate to dy/dx (recall dy/dx is just the gradient of the tangent line)?
(ii) P lies on the curve. Using (i), can you set up an equation to find the coordinates of P?
(Again, it's a good idea to have some descriptive words like "since the tangent at P is parallel to the y-axis, we have 'blah blah blah and calculations'")

On the mark scheme they split up the gradient for example to find P they use 4x-16y=0. I dont understand why they do this
Reply 4
Original post by toxicgamage56
Do you happen to have an answer as to what k is? If my long-winded method gives the correct answer, then I'll explain what I did, but I don't know if it's even right.


K=9/4
Original post by Matheen1
K=9/4

Yeah my answer's definitely not correct then. Since P and Q were on tangents that crossed the axis, I assumed you'd made the gradient from part a equal to zero since the axis are just straight lines. I saw your other response, and I don't get why that'd mean 4x - 16y = 0 though because I'd assume the denominator gets cancelled out when you cross multiply. Sorry I couldn't help.
Reply 6
Original post by Matheen1
On the mark scheme they split up the gradient for example to find P they use 4x-16y=0. I dont understand why they do this

Maybe two mini questions.
(i) The tangent line is parallel to the y-axis, so a vertical line. What is a gradient of a vertical line, loosely speaking?
(ii) We know dy/dx has to "equal to this value". When does this happen?
(edited 12 months ago)
Reply 7
Original post by toxicgamage56
Do you happen to have an answer as to what k is? If my long-winded method gives the correct answer, then I'll explain what I did, but I don't know if it's even right.


You must not give a full solutions as that will break the rules. The poster above has given a hint.
Reply 8
Original post by Matheen1
On the mark scheme they split up the gradient for example to find P they use 4x-16y=0. I dont understand why they do this


This is because the gradient is parallel to the y-axis ie infinite.
Reply 9
Original post by Muttley79
This is because the gradient is parallel to the y-axis ie infinite.


Ok thanks. But why do these only choose the denominator of the gradient?
Reply 10
Original post by Matheen1
Ok thanks. But why do these only choose the denominator of the gradient?


Anything/0 is infinte - the denominator comes into play for the other point.
Reply 11
Original post by Muttley79
Anything/0 is infinte - the denominator comes into play for the other point.

What other point?
Original post by Matheen1
What other point?

Flat points have a gradient of 0 whilst vertical points have a gradient of infinity. I just realised where I made my mistake. Since P is parallel to the y-axis where the gradient is vertical and equal to infinity, you'd use 4x - 16 y = 0 as anything/0 is infinity. If you put that into your calculator, it'll come up as a maths error. Then you'd just do the opposite for point Q where the gradient is not infinity, but 0 so 4x - 4y - 9 = 0. I'll try this out, and see if I get the answer that the mark scheme says.
Original post by Muttley79
You must not give a full solutions as that will break the rules. The poster above has given a hint.

Yeah, I'm aware. I was going to posit a suitable method if my answer was correct, which it wasn't.
Reply 14
Original post by Matheen1
What other point?

You need P and Q to get the distance between them.
Original post by Muttley79
You need P and Q to get the distance between them.

When you make the denominator of the derivative 0, you get 4x = 16y which is easy enough to sub into the curve's original equation to get a coordinate for P. However, I'm struggling to apply the same method for Q by making 4x - 4y - 9 equal 0 since putting this into the curve's original equation is difficult, and seems to give me impossible/convoluted values for the coordinates of Q. Is there something I'm doing wrong?
Reply 16
Original post by toxicgamage56
When you make the denominator of the derivative 0, you get 4x = 16y which is easy enough to sub into the curve's original equation to get a coordinate for P. However, I'm struggling to apply the same method for Q by making 4x - 4y - 9 equal 0 since putting this into the curve's original equation is difficult, and seems to give me impossible/convoluted values for the coordinates of Q. Is there something I'm doing wrong?

There are 8 marks - can you post your working?
Original post by Muttley79
There are 8 marks - can you post your working?

I've just been scribbling values I'm getting on paper whilst using an online calculator (lost mine so I need to order a new one).

For Q:

4x - 4y - 9 = 0
4x = 4y + 9
Original equation of curve: 4xy = 2x^2 + 8y^2 - 36y
(4y + 9)y = 2(4y+9/4)^2 + 8y^2 - 9(4y+9/4)
4y^2 + 9y = (16y^2 + 81)/8 + 8y^2 - (36y + 81)/4
I multiplied everything by 8 here to get rid of the fractions
32y^2 + 72y = 16y^2 + 81 + 64y^2 - (72y + 162)
Collected like terms
48y^2 - 144y - 81 = 0
This is the quadratic equation that I'm left with, and it gives me wacky coordinates.

When applying the same process for P, I just get (6, 1.5) which is a normal set of coordinates, so I'm unsure why the same doesn't work for Q.
Reply 18
Original post by toxicgamage56
I've just been scribbling values I'm getting on paper whilst using an online calculator (lost mine so I need to order a new one).

For Q:

4x - 4y - 9 = 0
4x = 4y + 9
Original equation of curve: 4xy = 2x^2 + 8y^2 - 36y
(4y + 9)y = 2(4y+9/4)^2 + 8y^2 - 9(4y+9/4)
4y^2 + 9y = (16y^2 + 81)/8 + 8y^2 - (36y + 81)/4
I multiplied everything by 8 here to get rid of the fractions
32y^2 + 72y = 16y^2 + 81 + 64y^2 - (72y + 162)
Collected like terms
48y^2 - 144y - 81 = 0
This is the quadratic equation that I'm left with, and it gives me wacky coordinates.

When applying the same process for P, I just get (6, 1.5) which is a normal set of coordinates, so I'm unsure why the same doesn't work for Q.

Id have noted that part of the expression is
4y^2 - 4xy = y(4y-4x) = -9y
which cuts down on work/errors. Then ...

But youve a few typo errors in the above.
(edited 12 months ago)
Original post by mqb2766
Id have noted that part of the expression is
4y^2 - 4xy = y(4y-4x) = -9y
which cuts down on work/errors. Then ...

Nevermind, I found out where I made my mistake. I foolishly thought that (4y+9)^2 was equal to 16y^2 + 81 for whatever reason, instead of correctly expanding the double bracket. The method is correct, and the answer now checks out for me. Thanks for the pointer though.

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