Help needed with parameters question!! (1 Viewer)

[ssukarieh1]

Hey guys, question 21 is the one i need help with. Its been bugging me for a while im sure im doing something silly but after numerous attempts all i seem to get is that (p+q)^2=4pq. Help is much appreciated!

View attachment 31672

 

[Squar3root]

I get (p+q)^2=4pq as well

 

[ssukarieh1]

Do you think it is a fault within the question? I doubt it as the question comes out of Cambridge...

 

[BLIT2014]

Do you think it is a fault within the question? I doubt it as the question comes out of Cambridge...

No textbook is 'perfect'.

 

[ssukarieh1]

Oh ok. Thanks for your help. In return i will complete your survey

 

[enigma_1]

I get (p+q)^2=4pq as well

Me too :/

I think the answers are wrong

 

[BLIT2014]

Oh ok. Thanks for your help. In return i will complete your survey

Thank you

 

[Squar3root]

Do you think it is a fault within the question? I doubt it as the question comes out of Cambridge...

so?

No textbook is 'perfect'.

this

even terry lee has mistakes in his book

Me too :/

I think the answers are wrong

yeah :/

Thank you

I'll do your survey too

 

[ssukarieh1]

OK, i was just implying that the mistakes in cambridge dont show up so often ! But if you all got that then it is obviously wrong! Thanks to all for your help !

 

[FrankXie]

Hey guys, question 21 is the one i need help with. Its been bugging me for a while im sure im doing something silly but after numerous attempts all i seem to get is that (p+q)^2=4pq. Help is much appreciated!

View attachment 31672

You guys are all wrong! If (p+q)^2=4pq were true, this means (p+q)^2-4pq=0, which implies (p-q)^2=0 and thus p=q, and this is a contradiction.

And indeed, I can show (p+q)^2=8pq, as desired.

 

[FrankXie]

Make sure you guys understood the question? it says the line PQ is a tangent to x^2=2ay, but NOT x^2=4ay

 

[ssukarieh1]

Could u please post up your solution?

 

[FrankXie]

Well, if a line is a tangent to the parabola x^2=2ay ( again NOT x^2=4ay), what condition must be satisfied?
If you can not solve it by tomorrow, i'll post my solution then.
The reason I thought u misunderstood that line PQ is a tangent to x^2=4ay is that if (p+q)^2=4pq then indeed PQ is a tangent to x^2=4ay. So you guys just need to adjust your solution using x^2=2ay instead of x^2=4ay.

 

[enigma_1]

Sorry I just realised what Frank said and yes the answer does come out. I reckon we all misread the question and it says x^2=2ay and not 4ay.
Basically solve the tangent and the curve simultaneously and you get a quadratic in terms of "x", and then do something with the discriminant ~~~~~

(hint: if it's a tangent think about how many solutions there would be to the quadratic equation, and use the discriminant accordingly)

 

[ssukarieh1]

Oh yes I see, I got the answer! Thanks again frank and enigma for your help!! It is much appreciated.