Prelim 2016 Maths Help Thread (1 Viewer)

InteGrand · Jul 9, 2016

Rathin said:
Just confirming something
Is the answer for sqrt((3x+2(x-3))<|x+3|, -1<x<15/2?

$\noindent We have $\sqrt{3x+2\left(x-3\right)} = \sqrt{5x -6}$. So in addition to solving the inequation, we need to also make sure $5x-6\geq 0$, so a condition on $x$ is that $x \geq \frac{6}{5}$.$

InteGrand · Jul 9, 2016

Why did they write it as 3x + 2(x-3)? Why not just immediately give it as 5x – 6? Was it a typo (should have been a quadratic term)?

Green Yoda · Jul 9, 2016

I am so sorry I made a typo and didn't include a bracket its
sqrt((3x+2)(x-3))<|x+3|

HeroicPandas · Jul 9, 2016

Rathin said:
I am so sorry I made a typo and didn't include a bracket its
sqrt((3x+2)(x-3))<|x+3|

x = 0 doesn't work. You are missing something in your solution (-1<x<15/2). Apply what InteGrand said (post #922) and you will get the answer.

Green Yoda · Jul 10, 2016

Oh I see...when ever we are solving for x in a square root we must make sure that the things inside the square root >0 yeah?
So the other answers would be x<-2/3 or x>3 yeah?

leehuan · Jul 10, 2016

Rathin said:
I am so sorry I made a typo and didn't include a bracket its
sqrt((3x+2)(x-3))<|x+3|

$\text{Initial restriction on }x:\\ \begin{align*} (3x+2)(x-3) &\ge 0 \\ \implies x\le -\frac{2}{3},\, & \, x \ge 3 \end{align*}$

$\text{Solving the inequality:}\\ \begin{align*}\sqrt{(3x+2)(x-3)}& < |x+3|\\ \implies 3x^2-7x-6 &< x^2+6x+9\\ \implies 2x^2 - 13x - 15 &< 0 \\ \implies (x+1)(2x-15) &< 0\\ \implies -1 < x& < \frac{15}{2}\end{align*}$

$\text{Combining cases}\\ -1 < x \le -\frac{2}{3}\\3 \le x < \frac{15}{2}$

Green Yoda · Jul 17, 2016

Find the equation of the parabola with: like x=2 as the axis, vertex(2,5), cuts y-axis at y=9
How do we find the focal length/latus rectum to find the value of a?
So far this is my equation (x-2)^2=+a(y-5)

davidgoes4wce · Jul 17, 2016

Rathin said:
Find the equation of the parabola with: like x=2 as the axis, vertex(2,5), cuts y-axis at y=9
How do we find the focal length/latus rectum to find the value of a?
So far this is my equation (x-2)^2=+a(y-5)

Do you have an answer it , just want to compare my answer what is right.

InteGrand · Jul 17, 2016

Rathin said:
Find the equation of the parabola with: like x=2 as the axis, vertex(2,5), cuts y-axis at y=9
How do we find the focal length/latus rectum to find the value of a?
So far this is my equation (x-2)^2=+a(y-5)

$\noindent The equation is of the form $\left(x-2\right)^2 = 4a \left(y-5\right)$, where $|a|$ is the focal length (the sign of $a$ here will dictate which way the parabola faces). To find $a$, sub. in $x=0$ and $y=9$, since the curve cuts the $y$-axis here. Once you've found $a$, you can draw a diagram to help see what to do to find the latus rectum.$

davidgoes4wce · Jul 17, 2016

InteGrand said:
$\noindent The equation is of the form $\left(x-2\right)^2 = 4a \left(y-5\right)$, where $|a|$ is the focal length (the sign of $a$ here will dictate which way the parabola faces). To find $a$, sub. in $x=0$ and $y=9$, since the curve cuts the $y$-axis here. Once you've found $a$, you can draw a diagram to help see what to do to find the latus rectum.$

Just curious to know what you call that 'a' constant? Do you call it the 'latus rectum'?

InteGrand · Jul 17, 2016

davidgoes4wce said:
Just curious to know what you call that 'a' constant? Do you call it the 'latus rectum'?

$\noindent That $a$ is just the focal length (in absolute value). The latus rectum is defined here, along with a picture (the blue chord in the picture in this link):$

http://mathworld.wolfram.com/LatusRectum.html .

Green Yoda · Jul 17, 2016

InteGrand said:
$\noindent The equation is of the form $\left(x-2\right)^2 = 4a \left(y-5\right)$, where $|a|$ is the focal length (the sign of $a$ here will dictate which way the parabola faces). To find $a$, sub. in $x=0$ and $y=9$, since the curve cuts the $y$-axis here. Once you've found $a$, you can draw a diagram to help see what to do to find the latus rectum.$

Thank you

Green Yoda · Jul 23, 2016

Find the locus of the locus p if the distance of p from line y=-5 is three quarters of its distance from the line x=2.
I did 3/4(|y+5|)=|x-2|

but looking at the answer it wants me to do 3/4(|x-2|)=|y+5|? Why?

InteGrand · Jul 23, 2016

Rathin said:
Find the locus of the locus p if the distance of p from line y=-5 is three quarters of its distance from the line x=2.
I did 3/4(|y+5|)=|x-2|

but looking at the answer it wants me to do 3/4(|x-2|=|y+5|? Why?

$\noindent The distance of the point $P\left(x,y\right)$ to the line $y=-5$ is $\left|y+5\right|$. Its distance to the line $x=2$ is $\left|x-2\right|$. So we need $\text{distance to }(y=-5) = \frac{3}{4}\text{ distance to }(x=2)\Longleftrightarrow \left|y+5\right| = \frac{3}{4}\left|x-2\right|$.$

Green Yoda · Jul 23, 2016

doesn't it say the distance of p to y=-5 is 3/4th than the distance from x=2?

InteGrand · Jul 23, 2016

Rathin said:
doesn't it say the distance of p to y=-5 is 3/4th than the distance from x=2?

Yeah, so we multiply its distance from x = 2 by 3/4 to get its distance from y = -5. E.g. If |x – 2| was 8, then |y + 5| would be 6, which is (3/4)*8 = (3/4)*|x – 2|.

Green Yoda · Jul 23, 2016

InteGrand said:
Yeah, so we multiply its distance from x = 2 by 3/4 to get its distance from y = -5. E.g. If |x – 2| was 8, then |y + 5| would be 6, which is (3/4)*8 = (3/4)*|x – 2|.

oh so the x=2 os 3/4th the distance from p when compared to p and y=-5?
Yeah the wording really confused me.

Green Yoda · Jul 23, 2016

find the locus of a point p that moves so that PA^2-PB^2=5
They didnt give coords of A and B..how do I do this?

Green Yoda · Jul 24, 2016

Orwell · Jul 28, 2016

Help lads. Need help with both.

Make a Difference – Donate to Bored of Studies!

Prelim 2016 Maths Help Thread (1 Viewer)

Well-Known Member

Well-Known Member

Hi Φ

Heroic!

Hi Φ

Well-Known Member

Hi Φ

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Hi Φ

Hi Φ

Well-Known Member

Hi Φ

Well-Known Member

Hi Φ

Hi Φ

Hi Φ

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)