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C2 - Logarithms

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Original post by B_9710
If there were another root of f, x>3 then there would have to be a turning point in the range (3,∞) (3,\infty ) .


Sorry I don't understand what you're saying :s-smilie:.
If there was a larger root of f(x) when x > 3 so a root which lies in a larger interval or not between [2 and 3] then what?
Original post by Chittesh14
Sorry I don't understand what you're saying :s-smilie:.
If there was a larger root of f(x) when x > 3 so a root which lies in a larger interval or not between [2 and 3] then what?


If there is a root that is greater than 3, then there must be a turning point at some point where x>3. Think about sketching the graph and what it would look like if a root did exist for x>3.
Original post by B_9710
If there is a root that is greater than 3, then there must be a turning point at some point where x>3. Think about sketching the graph and what it would look like if a root did exist for x>3.


Oh right, I understand. Thanks :smile: there must be a change of sign again... between another interval. Thanks!
Original post by B_9710
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Original post by RDKGames
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Original post by Zacken
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Original post by SeanFM
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Can anyone please confirm if my answers to the following questions are correct:

1. A point P(x, y) obeys a rule that the distance of P to the point (3, 0) is the same as the distance of P to the straight line x + 3 = 0. Prove that the locus of P has an equation of the form y2 = 4ax, stating the value of the constant a. a = 3

2. A point P(x, y) obeys a rule that the distance of P to the point (2√5, 0) is the same as the distance of P to the straight line x = -2√5. Prove that the locus of P has an equation of the form y2 = 4ax, stating the value of the constant a. a = 2√5
Original post by Chittesh14
Can anyone please confirm if my answers to the following questions are correct


Yeah, basic properties of parabolas.
Original post by Zacken
Yeah, basic properties of parabolas.


Yeah ik lol just wanted to see if answers were correct - thanks.
I am on Chapter 3 of FP1 btw :P ^ - parabola topics lol
Original post by B_9710
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Original post by RDKGames
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Original post by Zacken
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Original post by SeanFM
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Can someone help me on this question.

4. The parabola has parametric equation x = 6t^2, y = 12t. The focus to C is at the point S.

The points P and Q on the parabola are both at a distance 9 units away from the directrix of the parabola.

(e) Find the exact length PQ, giving your answer as a surd in its simplest form.
(f) Find the area of the triangle PQS, giving your answer in the form k√2, where k is an integer.

I keep getting 9√2 for (e) idk what I'm doing wrong lol.

Information:

The Cartesian equation of the parabola (C) is y2 = 24x.
The coordinates of the focus (S) is (6, 0) and the equation of the directrix is x + 6 = 0.
PS = 9 units
(edited 7 years ago)
Original post by Chittesh14
I keep getting 9√2 for (e) idk what I'm doing wrong lol.


Yes, well we can't tell what you're doing wrong without you including your working. Anywho, general strategy: a quick sketch should reveal that P and Q share the same x-coordinate (x=3) i.e: P and Q are vertically above each other, then from the cartesian equation you get that their y coordinates satisfy y^2 = 24(3) so the y coordinates of the two should be y=±24×3=±62y = \pm \sqrt{24\times 3} = \pm 6\sqrt{2} which makes total sense, since they are vertically above each other.

Then the distance between them is simply the difference in y coordinates 12212\sqrt{2}. Again, can't tell how you went wrong in getting your answer because you haven't posted your working, please try and do that in the future.
Original post by Zacken
Yes, well we can't tell what you're doing wrong without you including your working. Anywho, general strategy: a quick sketch should reveal that P and Q share the same x-coordinate (x=3) i.e: P and Q are vertically above each other, then from the cartesian equation you get that their y coordinates satisfy y^2 = 24(3) so the y coordinates of the two should be y=±24×3=±62y = \pm \sqrt{24\times 3} = \pm 6\sqrt{2} which makes total sense, since they are vertically above each other.

Then the distance between them is simply the difference in y coordinates 12212\sqrt{2}. Again, can't tell how you went wrong in getting your answer because you haven't posted your working, please try and do that in the future.


I always post my working out but my iPad is out of battery right now. Anyway, I don't get it lol can you draw the graph please. Why do P and Q have the x coordinate of 3?

Is it because -6 + 9 = 3. Nvm, I get it now. I kept doing (-6, 0) and trying to calculate it from a specific coordinate on the directrix which is where I went wrong. ****
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(edited 7 years ago)
Original post by Chittesh14
I always post my working out but my iPad is out of battery right now. Anyway, I don't get it lol can you draw the graph please. Why do P and Q have the x coordinate of 3?


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Sketch the parabola and the directrix. The only way two different points can have the same distance to it is if they are vertically above each other.

And the reason their x coordinate is 3 is because PS = 9, but that means the distance from P to the directrix is also 9 (defining property of parabola). So since the directrix is x=-6, then the distance OP is 9-6 = 3. Sketch a graph, all will be clear.
Original post by Zacken
Sketch the parabola and the directrix. The only way two different points can have the same distance to it is if they are vertically above each other.

And the reason their x coordinate is 3 is because PS = 9, but that means the distance from P to the directrix is also 9 (defining property of parabola). So since the directrix is x=-6, then the distance OP is 9-6 = 3. Sketch a graph, all will be clear.


Or if they're a reflection in the directrix lol :biggrin: - as in 9 units to the left of the directrix and 9 units to the right. But, we don't have the y value so that wouldn't work :P.

I know what you're saying about the parabola, but I was thinking something else - as in the y co-ordinate must be exactly the same but don't worry - it'll confuse you because I can't explain it lool.
Original post by Chittesh14
Or if they're a reflection in the directrix lol :biggrin: - as in 9 units to the left of the directrix and 9 units to the right. But, we don't have the y value so that wouldn't work :P.


Huh? That's not how a parabola works. If that was true, it'd have been a horizontal directrix, but it isn't, so it follows that they have the same x coordinate.
Original post by Zacken
Huh? That's not how a parabola works. If that was true, it'd have been a horizontal directrix, but it isn't, so it follows that they have the same x coordinate.


Ok nvm, I don't know lol.
Original post by B_9710
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Original post by RDKGames
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Original post by Zacken
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Original post by SeanFM
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Can someone help me on this question.

7 (b)
1. (e)

I'm stuck on 1. (e) and for 7 (b) I don't know how I got the answer, didn't understand it.

EDIT: Nvm, just understood 7 (b) lol...
(edited 7 years ago)
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Original post by Chittesh14
Can someone help me on this question.

7 (b)
1. (e)

I'm stuck on 1. (e) and for 7 (b) I don't know how I got the answer, didn't understand it.



Yeah, I'm not surprised you're getting it wrong. You're assuming that y2=7x⇒y=7xy^2 = 7x \Rightarrow y = \sqrt{7x} which is just plain wrong. Indeed, we have y=±7xy = \pm \sqrt{7x} and we are interested in y=−7xy = - \sqrt{7x} here (obviously so, from the co-ordinates, we want the negative branch of the parabola, sketch(!!!!!)). This gets you dydx∣(7,−7)=−12\frac{\mathrm{d}y}{\mathrm{d}x} \big|_{(7, -7)} = -\frac{1}{2} as required.

This is why covering C(?) and C4 parametric and implicit differentiation is heavily useful before doing FP1.
Original post by Chittesh14
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You might want to cover implicit differentiation from C4
Original post by Zacken
Yeah, I'm not surprised you're getting it wrong. You're assuming that y2=7x⇒y=7xy^2 = 7x \Rightarrow y = \sqrt{7x} which is just plain wrong. Indeed, we have y=±7xy = \pm \sqrt{7x} and we are interested in y=−7xy = - \sqrt{7x} here (obviously so, from the co-ordinates, we want the negative branch of the parabola, sketch(!!!!!)). This gets you dydx∣(7,−7)=−12\frac{\mathrm{d}y}{\mathrm{d}x} \big|_{(7, -7)} = -\frac{1}{2} as required.

This is why covering C(?) and C4 parametric and implicit differentiation is heavily useful before doing FP1.


I don't do C4 until next year and I don't know what C(?) you are talking about.
I keep forgetting because I'm rushing, I understand it anyway so I'm not worried lol. I am just rushing I think, might have to sketch now.


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Original post by Chittesh14
I don't do C4 until next year and I don't know what C(?) you are talking about.
I keep forgetting because I'm rushing, I understand it anyway so I'm not worried lol. I am just rushing I think, might have to sketch now.


I just think your method of differentiating is very prone to mistakes like these as well as misconceptions, whereas implicit differentiation or parametric differentiation provide much nicer methods.
Original post by Chittesh14
I don't do C4 until next year and I don't know what C(?) you are talking about.
I keep forgetting because I'm rushing, I understand it anyway so I'm not worried lol. I am just rushing I think, might have to sketch now.


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Why are you rushing learning and doing maths in the summer? It's not like you will be doing the exams next month >_>

Invest more time before you work on being faster
Original post by RDKGames
Why are you rushing learning and doing maths in the summer? It's not like you will be doing the exams next month >_>

Invest more time before you work on being faster


I don't even know if I'm rushing. But, I finish these questions in like 2 minutes so I'm fast and maybe I'm just going a bit too fast.


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