1. ## Rectanglar hyperbola question

Hi
can someone show me how would you find the equation express in the form $\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and asymptotes of the following equations:

1)$\displaystyle x^2-4y^2-4x-8y-16=0$
i got $\displaystyle \frac{(x-2)^2}{4}-(y-1)^2=6$

2)$\displaystyle 4x^2-8x-y^2+2y=0$
This is what i have done:
$\displaystyle 4(x^2-2x)-(y-2y)=0$

$\displaystyle (x-1)^2-\frac{(y-1)^2}{4}=0$

answer is $\displaystyle \frac{4(x-1)^2}{3}-\frac{(y-1)^2}{3}=1$

P.S

2. Hello, Paymemoney!

Your algebra is off . . .

Express in the form $\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$
and find the asymptotes of the following equations:

$\displaystyle 1)\;x^2-4y^2-4x-8y-16\:=\:0$

i got: $\displaystyle \frac{(x-2)^2}{4}-(y-1)^2\:=\:6$ . . . . no

We have: .$\displaystyle (x^2 - 4x) - 4(y^2 + 2y) \:=\:16$

Complete the square: .$\displaystyle (x^2 - 4x {\color{blue} \:+\: 4}) - 4(y^2 + 2y {\color{red}\:+\: 1}) \;=\;16 {\color{blue}\:+\: 4} {\color{red}\:-\: 4}$

And we have: .$\displaystyle (x-2)^2 - 4(y+1)^2 \:=\:16$

Divide by 16: .$\displaystyle \boxed{\frac{(x-2)^2}{16} - \frac{(y+1)^2}{4} \;=\;1}$

To find the asymptotes, set the equation equal to 0.

. . $\displaystyle \frac{(x-2)^2}{16} - \frac{(y+1)^2}{4} \:=\:0 \quad\Rightarrow\quad \frac{(y+1)^2}{4} \;=\;\frac{(x-2)^2}{16} \quad\Rightarrow\quad (y\;+\;1)^2 \;=\;\frac{(x-2)^2}{4}$

Take square roots: .$\displaystyle y + 1 \;=\;\pm\tfrac{1}{2}(x-2)$

Therefore: .$\displaystyle y \:=\:\pm\tfrac{1}{2}(x-2) + 1\;\;\Rightarrow\;\;\begin{Bmatrix}y \:=\:\frac{1}{2}x - 2 \\ y \:=\:-\frac{1}{2}x \end{Bmatrix}$

3. what about my second question?

4. Originally Posted by Paymemoney
Hi
can someone show me how would you find the equation express in the form $\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and asymptotes of the following equations:

1)$\displaystyle x^2-4y^2-4x-8y-16=0$
i got $\displaystyle \frac{(x-2)^2}{4}-(y-1)^2=6$

2)$\displaystyle 4x^2-8x-y^2+2y=0$
This is what i have done:
$\displaystyle 4(x^2-2x)-(y-2y)=0$

$\displaystyle (x-1)^2-\frac{(y-1)^2}{4}=0$

answer is $\displaystyle \frac{4(x-1)^2}{3}-\frac{(y-1)^2}{3}=1$

P.S
2) $\displaystyle 4x^2 - 8x - y^2 + 2y = 0$

$\displaystyle 4(x^2 - 2x) - (y^2 - 2y) = 0$

$\displaystyle 4(x^2 - 2x + 1) - (y^2 - 2y + 1) = 0 + 4 - 1$

$\displaystyle 4(x - 1)^2 - (y - 1)^2 = 3$

$\displaystyle \frac{4(x - 1)^2}{3} - \frac{(y - 1)^2}{3} = 1$.

Now to find the asymptotes, set the equation equal to 0.

So $\displaystyle \frac{4(x - 1)^2}{3} - \frac{(y - 1)^2}{3} = 0$

$\displaystyle \frac{(y - 1)^2}{3} = \frac{4(x - 1)^2}{3}$

$\displaystyle (y - 1)^2 = 4(x - 1)^2$

$\displaystyle y - 1 = \pm 2(x - 1)$

$\displaystyle y = 1 \pm 2(x - 1)$.

Therefore $\displaystyle y = 2x - 1$ or $\displaystyle y = -2x + 3$.

5. Originally Posted by Prove It
2)
$\displaystyle 4(x^2 - 2x + 1) - (y^2 - 2y + 1) = 0 + 4 - 1$
can you explain to me how did you get 0 + 4 - 1??

6. Originally Posted by Paymemoney
can you explain to me how did you get 0 + 4 - 1??
To complete the square in $\displaystyle x$ I had to add $\displaystyle 4$ to both sides and to complete the square in $\displaystyle y$ I had to subtract $\displaystyle 1$ from both sides.

7. can someone tell me where i have gone wrong in finding the x-intercepts for question 1.

$\displaystyle \frac{4(x-1)^2}{3}-{(0-1)^2}{3}=1$
$\displaystyle \frac{4(x-1)^2}{3}=1+\frac{1}{3}$
$\displaystyle 4(x-1)^2=\frac{4}{3}*3$
$\displaystyle (x-1)^2=\frac{12}{3}*\frac{1}{4}$
$\displaystyle x=\sqrt1+1$
so x=0 or x=2
i looked in the book's answers and it says $\displaystyle 1-\frac{\sqrt3}{2}$ and $\displaystyle 1+\frac{\sqrt3}{2}$

8. Originally Posted by Paymemoney
Hi
can someone show me how would you find the equation express in the form $\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and asymptotes of the following equations:

1)$\displaystyle x^2-4y^2-4x-8y-16=0$
i got $\displaystyle \frac{(x-2)^2}{4}-(y-1)^2=6$

2)$\displaystyle 4x^2-8x-y^2+2y=0$
This is what i have done:
$\displaystyle 4(x^2-2x)-(y-2y)=0$

$\displaystyle (x-1)^2-\frac{(y-1)^2}{4}=0$

answer is $\displaystyle \frac{4(x-1)^2}{3}-\frac{(y-1)^2}{3}=1$

P.S
The $\displaystyle x$ intercepts are where $\displaystyle y = 0$.

So $\displaystyle x^2-4y^2-4x-8y-16=0$

$\displaystyle x^2 - 4x - 16 = 0$

$\displaystyle x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-16)}}{2(1)}$

$\displaystyle = \frac{4 \pm \sqrt{16 + 64}}{2}$

$\displaystyle = \frac{4 \pm \sqrt{80}}{2}$

$\displaystyle = \frac{4 \pm 4\sqrt{5}}{2}$

$\displaystyle = 2 \pm 2\sqrt{5}$.

I do not agree with either answer...

9. oops its question 2 which i am having trouble finding the x-intercepts

10. $\displaystyle 4x^2 - 8x - y^2 + 2y = 0$

Let $\displaystyle y = 0$, then

$\displaystyle 4x^2 - 8x = 0$

$\displaystyle 4x(x - 2) = 0$

$\displaystyle 4x = 0$ or $\displaystyle x - 2 = 0$

$\displaystyle x = 0$ or $\displaystyle x = 2$.