Find the asymptotes of hyperbola 4x^2 + 8x − y^2 + 4y − 16 = 0Bring this equation to the standard form: Please I wanna understad some of the details.(4x^2 + 8x) − (y^2 − 4y) − 16 = 0,when...

Find the asymptotes of hyperbola 4x^2 + 8x − y^2 + 4y − 16 = 0
Bring this equation to the standard form:

Please I wanna understad some of the details.

(4x^2 + 8x) − (y^2 − 4y) − 16 = 0,
when we put parentheses around the y terms,
we should switch the signs inside,
because of the ‘ − ’ in front
4(x2 + 2x) − (y2 − 4y) − 16 = 0,
4(x2 + 2x + 1 − 1) − (y2 − 4y + 4 − 4) − 16 = 0,

My question is why (+1 and -1) and (+4 and -4) Please I wanna know why?

4((x + 1)^2 − 1)−((y − 2)^2 − 4)− 16 = 0, I know how this happend!!
4(x + 1)^2 − 4 − (y − 2)^2 + 4 − 16 = 0,
4(x + 1)^2 − (y − 2)^2 − 16 = 0,
4(x + 1)^2 − (y − 2)^2 = 16.

Now, using this equation of the hyperbola, we write the equation of the asymptotes:
4(x + 1)^2 − (y − 2)^2 = 0,

which we have to solve for y:
4(x + 1)^2 = (y − 2)^2,
extract the square root from both sides
±2(x + 1) = (y − 2),
this results in two equations:

−2(x + 1) = (y − 2), and 2(x + 1) = (y − 2),
−2x − 2 = y − 2, and 2x + 2 = y − 2,
−2x = y, and 2x + 4 = y.
These are the separate equations of each of the asymptotes. Answer: y = −2x, y = 2x + 4

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should remember what is the standard form of equation of hyperbola such that:

`((x-h)^2)/(a^2) - ((y-k)^2)/(b^2) = 1`

You need to group the terms that contain x alone and the terms that contain y alone such that:

`(4x^2 + 8x)- (y^2 - 4y) - 16 = 0`

You need to complete the squares such that:

`4(x^2 + 2x + 1) - (y^2 - 4y + 4) + 8 - 16 = 0`

`4(x+1)^2 - (y-2)^2 = 8`

You need to divide by 8 to make the right side to become 1 such that:

`((x+1)^2)/2 - ((y-2)^2)/8 = 1`

You should know that the equation of asymptote of hyperbola is given by equation `y = mx + c`

m represents the slope of asymptote

`m = b/a`

Since `b = 2sqrt2`  and `a = sqrt2 => m = 2sqrt2/sqrt2 =>m = 2`

`y = +-2(x+1) + 2`

Hence, evaluating the asymptotes of the given hyperbola yields `y = +-2(x+1) + 2.`

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