help! i'm going insane with these questions (1 Viewer)

[Ellie999]

i'm such a failure, i can't even solve these simple questions, and trial is on tomorrow. *slash wrist* there's big fat chance of me phailing...omg....someone please explain how to solve these questions:
1. if 3 is a root of the equation x^2+kx-6=0, find the value of k.

2.express 4x^2 in the form A(x-1)^2+B(x-1)+C.

3.express 0.37(infinite numbers-it keeps going as 0.3737373737...) as a geometric series and find its limiting sum

4.find the equation of the parabola with focus at (0,3) and directrix y=-3
ty in advance

p.s. i forgot to post 1 more question...
how do you show that 2x^2-3x+6 is positive for all values of x?

 

[independantz]

4.find the equation of the parabola with focus at (0,3) and directrix y=-3
ty in advance

x^2=4ay (a=3)
X^2=12y

Not entirely sure if that's correct but seems right.

 

[Ellie999]

x^2=4ay (a=3)
X^2=12y

Not entirely sure if that's correct but seems right.

its right but im not following what you typed im a troll, beat me.

 

[independantz]

Well since the focus is at (0,3) and the directrix is y=-3 it means the parabola takes the cartesian equation x^2=4ay, you know how there's 4 of them depending on how the parabola is placed on the graph.

and a=3 because the focus is (a) units up from the number plane i.e the typical point of the Focus is (0,a), hope that helps lol.

 

[ssglain]

I will provide hints so you can try to solve these questions on your own. The solutions are under the spoilers.

1. If 3 is a root of the equation x² + kx-6=0, find the value of k.
Hint: In plain English, x = 3 satisfies the given equation.

Spoiler

This means that when you substitute 3 into x² + kx-6, your answer will be 0.
So, (3)² + 3k - 6 = 0
9 + 3k - 6 = 0
3k + 3 = 0
.: k = -1

2. Express 4x² in the form A(x - 1)² + B(x - 1) + C.
Hint: Expand A(x - 1)² + B(x - 1) + C and compare its coefficients with 4x².

Spoiler

A(x - 1)² + B(x - 1) + C
= A(x² - 2x + 1) + B(x - 1) + C
= Ax² - 2Ax + A + Bx - B + C
= Ax² + (B - 2A)x + (A - B + C) [collect like terms]
Let 4x² + 0x + 0 = Ax² + (B - 2A)x + (A - B + C)
Then by comparing coefficients of like terms:
A = 4 ...(1)
B - 2A = 0 ...(2)
A - B + C = 0 ...(3)

Substitute A = 4 into (2):
B - 2(4) = 0
.: B = 8

Substitute A = 4, B = 8 into (3):
4 - 8 + C = 0
.: C = 4

Hence, 4x² = 4(x - 1)² + 8(x - 1) + 4

3. Express 0.3737373737... as a geometric series and find its limiting sum.

Hint: Consider 0.3737373737... as the sum of a geometric sequence

Spoiler

0.3737373737... = 0.37 + 0.0037 + 0.000037 + ...

Clearly a = 0.37 and r = 0.001

Use the limiting sum formula S(∞) = a/(1 - r)
.: S(∞) = 0.37/(1 - 0.01) = 37/99

4. Find the equation of the parabola with focus at (0, 3) and directrix y = -3.

Hint: The standard equation of a parabola with vertex at (0, 0) is x² = 4ay, where a = focal length. This means that you need to deduce the focal length of this parabola from the information given. Recall that the perpendicular distance from the focus to the directrix = twice the focal length.

The solution is as shown by independentanz.

5. Show that 2x² - 3x + 6 is positive for all values of x.
There are a number of ways to do this. I will show two of them.

i) By calculus:
Find the turing point of y = 2x² - 3x + 6:
y' = 4x - 3 = 0 [for stat. pts.]
.: x = 3/4, y = 39/8

Determine the nature of (3/4, 39/8):
y" = 4 > 0
When x = 3/4, y" > 0
.: (3/4, 39/8) is a minimum turning pt.

.: The parabola is always above the x-axis since the minimum turning pt lies above the x-axis.
.: 2x² - 3x + 6 is positive for all values of x.

ii) By considering discriminant:
∆ = (-3)² - 4(2)(6) = -39 < 0
Coefficient of x² > 0
.: y = 2x² - 3x + 6 is a positive definite
.: 2x² - 3x + 6 is positive for all values of x.

 

[Ellie999]

to independantz...i'm such a clutz! thank you, i get it now (finally)

to ssglain: you are amazing. seriously, you are so awesome at maths, esp. when you are explaining solutions! *whispers* i wish that you finished HSC last year so i could get your tuition! oh well

 

[crazysambo]

u have trials tomorrow. damn! good luck, ours start monday and first week is all english and SOR

 

[Ellie999]

hi pplz..inno that this is an old thread but i didnt want to waste the space by creating anotha one... i now have probs with LOCUS! and im using cambridge! yikes.....i spent at least an hour on each and im not getting anywhere...
so here are the evil questions...:

1.by using the perpendicular distance formula, find the locus of a point P (x,y) which is equidistant from teh lines 3x+4y=36 and 4x+3y=24

2.use the perpendicular dist. formula to find the distance of the point p(x,y) from each of the sides of the triangle. hence find the locus of the point p(x,y) which moves so that the sum of the squares of the distances from the sides of the triangle is 9.

3.a)the equation of a parabola is of the form y=kx^2. if the line 8x-y-4=0 is a tangent to the parabola, find the value of k.
b)a parabola with vertical axis has its vertex at the origin. find the equation of the parabola if the line 12x-4y+3=o is a tangent

...sorry for asking these many questions

 

[aaduckie]

Hi Ellie,

I'm pretty sure you haven't given all the information for question 2. What triangle?

For question 1 you use the perp distance formula (Am+Bn+C/(A^2+B^2)^0.5) (sorry it looks messy in text form)

m = x and n = y (given by the point p (x,y))

Now, since the distance between the point p and the two lines will always be equal, you make the two perp distance formulas equal to each other and then simplify. This should be really straight forward once you know what the question is asking you to do. The formula is the locus (or path of points) which always meets the criteria of being the same distance from one line as it is from the other. Note: Sometimes the locus is a parabola, sometimes a straight line and sometimes a different kind of curve.

I hope that helps with question 1 anyway, it might also give you some hints as to the other questions too.

Regards,
Amanda

 

[Forbidden.]

x^2=4ay (a=3)
X^2=12y

Not entirely sure if that's correct but seems right.

Well since the focus is at (0,3) and the directrix is y=-3 it means the parabola takes the cartesian equation x^2=4ay, you know how there's 4 of them depending on how the parabola is placed on the graph.

and a=3 because the focus is (a) units up from the number plane i.e the typical point of the Focus is (0,a), hope that helps lol.

I imagine a virtual Cartesian plane of a parabola and I agree about x²=12y

And ... you do exactly the same subs as me except you have SOR as your extra unit(s).