Mathematics Question Thread

[RuiAce]

Hi!

I'm having trouble with a line in the 2017 HSC Sample Answers for Q16 c) i) if anyone could please help me with it  - Attached

They get 'then BD = DE (equal intercept)', but I am unsure how they got this?

Much appreciated!! 

[michael.shin]

qs probability

[RuiAce]

qs probability

[alisoneom]

Hey guys! I need help with both parts of this HSC 2012 question attached. I’m most confused about the solution, which states that “The circle is a tangent if there is precisely one solution, so the discriminant has to vanish”, and then equates the discriminant to 0.

1) I don’t understand what that statement means ((
2) I’m not sure why they equated the discriminant to 0 (implying that there is only one solution) - shouldn’t there be two solutions, since the circle touches the parabola twice?
(It would also be great to have more guided answers to the remainder of the question)

Thank you in advance and I’d appreciate anyone’s help!

[alisoneom]

Another thing - I don't know how likely is it for this to come up in tomorrow's exam, but how do I differentiate from first principles...?

[fun_jirachi]

Hey there!

Just two things to start off with:
1. Please don't double post! There's an edit button on your original post if you want to append anything.
2. As for your first question, Rui already addressed this on the previous page! You might want to check it out.

Differentiating from first principles is essentially what you started out with when before you did all your product rule and your chain rule and your f'(x)=nx^n-1 jazz. Remember from the definition of the function that f(x)= whatever the function is is essentially the value of the function when you sub in x for x. A similar thing happens for f(x+h), sub in (x+h) every time you see an x, that's a great way to think about it. An example would be f(x)=x², you get that f(x+h)=(x+h)².

Differentiating from first principles is usually easy from here. You ideally want to factorise out the h and try to remove the f(x) by relating it to f(x+h) in some shape or form, so you get an expression independent of h (in the denominator) since the denominator can't equal zero.

If you ave any more questions, feel free to ask! Hope I helped

[Opengangs]

Hey guys! I need help with both parts of this HSC 2012 question attached. I’m most confused about the solution, which states that “The circle is a tangent if there is precisely one solution, so the discriminant has to vanish”, and then equates the discriminant to 0.

1) I don’t understand what that statement means ((
2) I’m not sure why they equated the discriminant to 0 (implying that there is only one solution) - shouldn’t there be two solutions, since the circle touches the parabola twice?
(It would also be great to have more guided answers to the remainder of the question)

Thank you in advance and I’d appreciate anyone’s help!

Hey, alisoneom!

(I've avoided looking at the solution to provide a full explanation into how to approach this type of question!)
First, notice that the parabola and the circle touch at some point. So we solve the two curves simultaneously. Substituting \(y = x^2\) into the circle, we get:

\[ y + (y - c)^2 = r^2.\]

Upon expanding, you'll end up with a quadratic in terms of \(y\). The question tells us that the point of intersection is symmetric for \(y\). So, while we do have two points of intersection, these points of intersection occur on the same \(y\) value! This tells us that you have one distinct root. And hence, you have the discriminant being zero, which answers your first and second question!

Setting the discriminant equal to 0 will give you the required result!

[jamesrandom]

Hey can someone help me out with this question. Have no clue where im going wrong....

[RuiAce]

Hey can someone help me out with this question. Have no clue where im going wrong....

Or if you chose the product rule approach to differentiate it you'd obtain \( \frac{dy}{dx} = 2(x+3)^2 + 4x(x+3) \), but expanding that out still gives \(6x^2+24x+18\).

[jamesrandom]

Can someone just quickly explain why x > +- 3 becomes x > 3 and x < -3. Why does the inequality sign switch.

[RuiAce]

Can someone just quickly explain why x > +- 3 becomes x > 3 and x < -3. Why does the inequality sign switch.

It doesn't.

Are you solving a quadratic inequality? Because note that for quadratic inequalities such as \(x^2-9 > 0\), you should not be treating the same way as though you were solving linear inequalities.

[Bilbo]

Hi, found the lecture super helpful, will these come out each month.

[jamonwindeyer]

Hi, found the lecture super helpful, will these come out each month.

Welcome to the forums!! Our Maths lectures typically happen in the holiday periods, so every few months

[jamesrandom]

Plz help this locus chapter is doing my head in

[S200]

Plz help this locus chapter is doing my head in

Question 8

\(Eq_1: y=2x-4\) (re-arranged)
\(Eq_2: y=\frac{1}{4}x^2\) (also re-arranged)
So, we know that the derivative at a point will equal the gradient of the tangent at that point (\(2\)), so lets go ahead and do that.
\(2=2\times \frac{1}{4}x \quad \therefore \quad x=4\)
Then, sub this x-value back into either of the given equations to get the y-coordinate, and hence the full co-ordinates of the point.

Question 10

I had no clue what a focus or a latus rectum was before this question, so thanks for posting this question!

This gives a nifty formula to help here.

If you have the equation of a parabola in vertex form \(y=a(x−h)^2+k\), then the vertex is at \((h,k)\) and the focus is \((h,k+\frac{1}{4a})\).

So, we re-arrange into the "vertex form"...
\(x^2=-8y \quad \therefore \quad y=\frac{-1}{8}(x)^2\)
We can clearly see that this parabola has the vertex at \((0,0)\), because \(h\) and \(k\) are not present (i.e; They are zero).
So, we move on to the focus, using the formula given... \((h,k+\frac{1}{4a})\)
\(h=0\), so no worries there.
\((k+\frac{1}{4a}) = \left (0+\frac{1}{4\times \frac{-1}{8}}\right )\) which then gives us the \(y\) (or \(k\)) position of \(-2\)

Now, we just equate the parabola to the line of \(y=-2\) to get the x-points of the intersection. The length of the latus rectum can be found by taking the length of the semi-latus rectum (i.e; the positive x-coordinate of the intersection points you just found) and doubling it.

Did you try to work it out?

Yes

\(2=\frac {x^2} {8}\) as the negatives cancelled.
\(x=\sqrt {2\times 8}\)
\(x=\pm 4 \quad \therefore \quad 4\times 2=8\), so the Latus Rectum is equal to \(8\) units!

No

Go try it please, 'cause just following my answers most likely won't help unless you try similar questions yourself and develop your problem solving abilities (and that's what we're all about) 

How did you go about working them out?