1. ## Rectanglar hyperbola question

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

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

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

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

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

P.S

2. Hello, Paymemoney!

Your algebra is off . . .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

5. Originally Posted by Prove It
2)
$4(x^2 - 2x + 1) - (y^2 - 2y + 1) = 0 + 4 - 1$
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 $x$ I had to add $4$ to both sides and to complete the square in $y$ I had to subtract $1$ from both sides.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

I do not agree with either answer...

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

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

Let $y = 0$, then

$4x^2 - 8x = 0$

$4x(x - 2) = 0$

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

$x = 0$ or $x = 2$.