HSC Mathematics Marathon (2 Viewers)

[blackops23]

Can someone please answer my conics question on like page 2 or 3, so I can get some questions going too.

Here's the FULL SOLUTION, only doing this because this is the one of the two questions i've been able to do in the marathon =(, also I dunno, if Anchovies still wants to know how to do it, (assuming he doesn't...)

--------

New question:

P(x1, y1) is a point on the hyperbole x^2/a^2 - y^2/b^2 = 1 with foci S, S'

(a) Find the equation of the tangent to the hyperbole at P.
(b) The tangent meets the x-axis in T. Show that PS/PS' = TS/TS'

(a) Implicit diffrentiation:
2x/a^2 - 2yy'/b^2 = 0
y' = (2x/a^2)*(b^2/2y)
y' = [b^2(x)]/[a^2(y)]
y-y1 = [b^2(x)/a^2(y)]*(x-x1)
Multiply through a^2(y)
(a^2)(y^2) - a^2(y)(y1) = (b^2)(x^2) - (b^2)(x)(x1)
(b^2)(x)(x1) - (a^2)(y)(y1) = (b^2)(x^2) - (a^2)(y^2)
Divide by (a^2)(b^2)

Tangent: (x)(x1)/a^2 - (y)(y1)/b^2 = 1
---------------------------------------------------

(b) To find T, sub in y=0 in equation of tangent.

therefore T=(a^2/x1 , 0)

Now e = PS/PM = PS'/PM'

So (PS/PM)divided by(PS'/PM') = e/e = 1
therefore:
(PS/PS')*(PM'/PM)=1
(PS/PS') = PM/PM'
Now PM/PM' =
= [x1 - (a/e)]/[x1 + (a/e)]
= [e(x1)-a]/[e(x1) + a]

Now TS/TS' = [ae - (a^2/x1)] / [ae + (a^2/x1)]
= [ae(x1) - a^2]/[ae(x1) + a^2]
= [e(x1) - a]/[e(x1) + a]
= PS/PS'

On paper it looks a lot more simpler...

 

[UnrealAnchovies]

Here's the FULL SOLUTION, only doing this because this is the one of the two questions i've been able to do in the marathon =(, also I dunno, if Anchovies still wants to know how to do it, (assuming he doesn't...)



(a) Implicit diffrentiation:
2x/a^2 - 2yy'/b^2 = 0
y' = (2x/a^2)*(b^2/2y)
y' = [b^2(x)]/[a^2(y)]
y-y1 = [b^2(x)/a^2(y)]*(x-x1)
Multiply through a^2(y)
(a^2)(y^2) - a^2(y)(y1) = (b^2)(x^2) - (b^2)(x)(x1)
(b^2)(x)(x1) - (a^2)(y)(y1) = (b^2)(x^2) - (a^2)(y^2)
Divide by (a^2)(b^2)

Tangent: (x)(x1)/a^2 - (y)(y1)/b^2 = 1
---------------------------------------------------

(b) To find T, sub in y=0 in equation of tangent.

therefore T=(a^2/x1 , 0)

Now e = PS/PM = PS'/PM'

So (PS/PM)divided by(PS'/PM') = e/e = 1
therefore:
(PS/PS')*(PM'/PM)=1
(PS/PS') = PM/PM'
Now PM/PM' =
= [x1 - (a/e)]/[x1 + (a/e)]
= [e(x1)-a]/[e(x1) + a]

Now TS/TS' = [ae - (a^2/x1)] / [ae + (a^2/x1)]
= [ae(x1) - a^2]/[ae(x1) + a^2]
= [e(x1) - a]/[e(x1) + a]
= PS/PS'

On paper it looks a lot more simpler...

Is this the same method Terry Lee uses? Because I can't be bothered reading all that writing lol.

Good job though.

 

[blackops23]

Is this the same method Terry Lee uses? Because I can't be bothered reading all that writing lol.

Good job though.

I dunno if Terry Lee did this, but this is my own method I created for my exam - there was an almost similar question, just involving an ellipse instead of a hyperbola in my exam.

 

[UnrealAnchovies]

I dunno if Terry Lee did this, but this is my own method I created for my exam - there was an almost similar question, just involving an ellipse instead of a hyperbola in my exam.

I didn't need it answered, I just wanted to keep the thread going with non-complex number questions.

Interesting method though.

 

[Trebla]

Show that for all real x:
1 + x² > 2x

Hence deduce that:
e^x > 1 + x²

 

[jyu]

Show that for all real x:
1 + x² > 2x

Hence deduce that:
e^x > 1 + x²

Impossible. Just try x=-1.

 

[funnytomato]

Show that for all real x:
1 + x² > 2x

Hence deduce that:
e^x > 1 + x²

don't think it's a legit question. cos it's true if x is restricted to non-negative real numbers. and I haven't really seen why x can't be -ve

 

[deterministic]

Impossible. Just try x=-1.

+1. This highlights the flaw in assuming if f(x)>g(x) for all x, then the primitives of both functions share the sample property (ie.F(x)>G(x) for all x) due to differing constants of integration.

 

[cutemouse]

Here you said f(x) is a rule. The other post you said f(x) is a value. What is it?

f(x)=x^2+1 is a rule for the function f.

f(x) is the value of the function f at the point 'x'.

 

[jyu]

Lol, if you're going to be all smart using notation not used in the HSC then at least get it all right. f(x) isn't a function; it's a rule.

So you are now saying you were talking nonsense when you said 'f(x) isn't a function; it's a rule'.

 

[Trebla]

Woops, screwed up the question. Here it is again:

Show that for x > 0:
1 + x² > 2x

Hence deduce that for x > 0:
e^x > 1 + x²

 

[cutemouse]

So you are now saying you were talking nonsense when you said 'f(x) isn't a function; it's a rule'.

f(x)=x^2+1 is a rule for the function f.

f(x) is the value of the function f at the point 'x'.

Woops, screwed up the question. Here it is again:

Show that for x > 0:
1 + x² > 2x

Hence deduce that for x > 0:
e^x > 1 + x²

Isn't the first part true for all real values of x?

 

[Bored Of Fail 2]

Isn't the first part true for all real values of x?

yes but its quite obvious you use it in the second part, and since the second part is only for positive x they decided they might as well make the first part prove for x>0

 

[funnytomato]

some one should make a new thread about how to use LaTex since the old one isn't working and I haven't even found where it is yet,lol

 

[Trebla]

Isn't the first part true for all real values of x?

It still holds for x > 0 (which is necessary for the second part) which is a subset of the reals.

 

[nrlwinner]

I'll put up a question for now:

Show, by mathematical induction, that |z_1+z_2+⋯+z_n |≤|z_1 |+|z_2 |+⋯+|z_n |

 

[cutemouse]

Show, by mathematical induction, that |z_1+z_2+⋯+z_n |≤|z_1 |+|z_2 |+⋯+|z_n |

That's boring.

 

[deterministic]

^that and induction is not required for this question.

 

[cutemouse]

^that and induction is not required for this question.

Well, technically it is. But still it's boring.

 

[jyu]

I'll put up a question for now:

Show, by mathematical induction, that |z_1+z_2+⋯+z_n |≤|z_1 |+|z_2 |+⋯+|z_n |

I'll do it anyway:

|z_1|≤|z_1 | true
Assume |z_1+z_2+⋯+z_k |≤|z_1 |+|z_2 |+⋯+|z_k | true
.: |z_1+z_2+⋯+z_k+1 |≤|z_1+z_2+⋯+z_k |+|z_k+1|≤|z_1 |+|z_2 |+⋯+|z_k | +|z_k+1| true
.: |z_1+z_2+⋯+z_n |≤|z_1 |+|z_2 |+⋯+|z_n | true