Question

Help please thanks :DIf the y-axis is tangential to the curve px^2+y^2-2px+2qy+r=0, fin r in terms of q. If this term does not meet the x-axis, find the range of values of p in terms of q

Answer 1

At the point where y is tangential to the curve x = 0 right?

Answer 2

yup that part i got

Answer 3

i also tried equating discriminant and this but hm (thanks again)

Answer 4

- by the way - the last question about the parabola the value of a is -0.5

Answer 5

ACK WAIT WHAT i got 5..lemme check

Answer 6

no need to use the vertex form

Answer 7

i keyed in my answer and it's right. like i check with online grapher

Answer 8

what a = 5?

Answer 9

at least the intercepts are

Answer 10

a must be negative as the graph has a maximum

Answer 11

check the post

Answer 12

oh no. what is this. :( is it not 5. damn. yeah it has to be negative what am i doing. sorry i'lll go check. meanwhile you mind helping me with this qn?

Answer 13

omg i get it (the other qn) but i don't get why i'm wrong using the vertex method

Answer 14

I am trying I had an operation yesterday and i'm not at my best.

Answer 15

ack! Rest well. Don't need to help me haha just go and rest :D

Answer 16

Yes i think i will

Answer 17

:D

Answer 18

@phi (if you want to solve this one that is)

Answer 19

ok
we have
\[px^2 + y^2 -2px +2qy +r=0\]

we know that the y axis is tangent
which means y' at x=0 is infinity

differentiate
\[2px +2yy' -2p +2qy' =0 \\ px +y'(y+q) -2p =0\\at~x=0\\y' =\frac{2p}{y+q}\]

Answer 20

we want y' equal to infinity
so y=-q

so (0,-q) is a point on the graph,
substitute
\[ q^2 -2q^2 +r=0\\q^2=r\]

Answer 21

wait what is y'

Answer 22

y' is dy/dx

i just wrote y' because its a neater notation

Answer 23

making sense now?

Answer 24

for the second part,

we know when y=0, x is not zero

so sub y=0 in the equation to get
\[px^2 -2px +r=0\]

we know that r=q^2
\[px^2 -2px +q^2=0\]

Answer 25

i dont know where to go from here,
perhaps complete the square for x^2 terms?

Answer 26

sorry what's dy/dx?

Answer 27

the derivative of y with respect to x

Answer 28

y' gives the slope of the tangent to the line. The y-axis was the the tangent - a vertical line whose slope is infinite.

Answer 29

@welshfella does the \(q^2=r\) calculations look correct?
any idea how to proceed for the second part of the question?

Answer 30

yes . Looks correct.

No. I'm not sure how to proceed with part 2.

Answer 31

'If this term does not meet the x axis' is confusing Shouldn't it be 'If the curve'

Answer 32

yea, i just assumed thats what it meant

Answer 33

ok
we have
\[px^2 -2px +q^2=0\]

this equation should not have x=1 as root,
so we do the whole b^2-4ac thing ?

Answer 34

*x=0 as a root

Answer 35

i just get q not equal to zero...

Answer 36

why not use completing the square?

Answer 37

yea
\[(x-1)^2 = \frac{p-q^2}{p}\]

so
\[(0-1)^2 \neq \frac{p-q^2}{p}\]

Answer 38

so again, just q \(\neq\)0 ?

Answer 39

yea I see what you mean . I cant see any problem with that.

Answer 40

To be honest I always struggle a bit with this type of problem.

Answer 41

oh well.. i guess we'll leave it at this. @wcrmellissa2001 will have to verify.

Hope you recover quickly @welshfella , good luck!

Answer 42

yes thank you . I'm still not at full strength.

Answer 43

and yet you cant keep away from openstudy can you :P
it's an addiction

Answer 44

right!!

Answer 45

This curve is an ellipse

how about this

(x - 1)^2 = p - q^2
------
p

when x = 1 p - q^2
------- = 0
p

and the curve touches the x-axis at this point

p = q^2 at x = 1

any value of q^2 greater than p would make (x - 2)^2 negative

so for p > x^2 the curve will be lower than the x-axis

rhat's the range they want

Answer 46

https://www.desmos.com/calculator/7varuvy9np

Answer 47

THANKS GUYS :D

Answer 48

maybe my teacher can clarify

Answer 49

yes
i'm pretty sure the above is right though
q^2 cant be greater than p because a square cant be negative
and as you see from the graph when p = q^2 the curve just touches the axis

when p > q^2 the whole curve is below the x axis.

Answer 50

ahh i actually sort of get it. I'm so proud. REST WELL @WELSHFELLA!!

Answer 51

yes OK

I'm sure you've learned something today.

Answer 52

yes!! what was i thinking lol

if it never intersects the x axis then this
\[(x-1)^2 = \frac{p-q^2}{p}\]
should have no real solutions

nice one @welshfella

Answer 53

i thnik it's something to do with discriminants for that one

Answer 54

yea
for there to be 2 real roots D > 0
one root D = 0
and no roots (doesn't pas the x-axis ) D < 0.

Answer 55

lets try that

Answer 56

D = 4p(^2 - 4pq^2 < 0 for the graph not to touch the x axis

4p(p - q^2) < 0

this gives p < q^2


- the opposite to what i said earlier!!!

I'll draw some more graphs to confirm this

Answer 57

https://www.desmos.com/calculator/nr07exibs6


check this one out
p = 8 and q^2 = 4

so p> q^2 is wrong

I'll try one with p < q^2

Answer 58

ttps://www.desmos.com/calculator/wjlgertejj

Yes this is not touching the x axis

that confirms p < q^2

Answer 59

There must have been a flaw in my argument which showed p > q^2 but I cant see it.

You were right @wcrmelissa2001 - the discriminant was the way to go!

Answer 60

Yup haha just so happened I checked with my classmate before I saw this... method is same as yours! Thanks :D :D