Rectangular hyperbola SGS 2010 Q7a (1 Viewer)

[asianese]

(a) (i) (2 marks)

(ii) (4 marks)

(iii) Show that there can never be more than four normals drawn to the hyperbola from an arbitrary point in the plane. (1 mark)

I've done 1 (trivial). Struggling for ii and iii. Do we solve the point with the normal?? Leads to nothing...

Any help is appreciated.

 

[Carrotsticks]

(ii) Substitute the point (0,k) into the equation in part (i) and show that there are EXACTLY two solutions for the quartic in terms of t.

(iii) If we sub in any arbitrary point (x,y) into the equation of the normal, we will acquire a polynomial of degree 4 in terms of t. By the Fundamental Theorem of Algebra, this polynomial has AT MOST 4 solutions (4 values of t) and hence at most 4 normals can be drawn.

 

[asianese]

OMG YEAH!...I didn't think of t as a variable - but here it's parametric. Thanks!

I can't solve tk=c-c*t^4....any ideas?

 

[Carrotsticks]

OMG YEAH!...I didn't think of t as a variable - but here it's parametric. Thanks!

I can't solve tk=c-c*t^4....any ideas?

You're not meant to solve it. You just need to prove that it has exactly 2 solutions.

 

[Carrotsticks]

Oh and regarding the green part, it really should be said the other way around: "If a quartic has EXACTLY one turning point below the X axis and it is a positive quartic, then it has EXACTLY two solutions".

But of course that is not necessarily the case because you can easily have 3 turning points underneath the X axis and still have two solutions, which is why I didn't say "iff" and just said "if" instead.

Oh and incase you haven't spotted, I also didn't actually PROVE that it was below the X axis, I just made a logical assertion, but I doubt the markers wanted a rigorous proof as to why that is the case.

 

[asianese]

Oh wow thanks a lot carrot!!

 

[CcYann]

You're not meant to solve it. You just need to prove that it has exactly 2 solutions.

View attachment 25380

is there another way to prove ii)? I mean how can I prove it if I don't know the suspicion or I couldn't think of it?

 

[Carrotsticks]

is there another way to prove ii)? I mean how can I prove it if I don't know the suspicion or I couldn't think of it?

I suppose you could sub in the stationary point to P(t) again and show it's negative, but that would be pretty difficult I can imagine since K can be positive or negative.

 

[seanieg89]

Whilst I am pretty sure Carrot's reasoning works, I think it is actually a little tedious to prove that the suspicion is correct. The most anal of HSC markers may object to such a proof by pictures as it is indeed quite a jump from the related (and commonly used in 4U) argument: f(a)<0, f(b)>0, f continuous => f has root between a and b.

Here is the outline of an algebraic solution:

 

[seanieg89]

One more thing, i) rules out the possibility of the two real roots coinciding. (If the two real roots had the same sign then the product of roots would be positive). Hence the polynomial has exactly two distinct real roots.

 

[iSplicer]

You're not meant to solve it. You just need to prove that it has exactly 2 solutions.

View attachment 25380

Your solutions are.. beautiful. The colours, the neatness, the rigorous set notation... wow. Mirin' dedication!

 

[Sanjeet]

^Agreed!!