1)**Identify the equation of the hyperbola with its center at (3, -1), its vertex at (5, -1), and an asymptote of 2y = 3x – 11.**

A) 9(x - 3)^{2} - 4(y + 1)^{2} -36 = 0

B) 4(y - 2)^{2} - 9(y + 1)^{2} = 36

C) (y + 3)^{2} + (x -1)^{2} + 36 = 0

D) 4(y +1)^{2} - 9(x -3)^{2} = 36

2) **Identify the equation of the ellipse that passes through (4, 2) and has foci at (1, -1) and (1, 5).**

A) (y + 2)^{2} + 2(x + 1)^{2} - 18 = 0

B) (y -2)^{2} + 2(x -1)^{2} = 18

C) (y -2)^{2} - 2(x - 1)^{2} = 18

D) 2(y + 2)^{2} – (x + 1)^{2} = 18

Guest Nov 11, 2014

#2**+5 **

I can see straight of from the formula that the centre will be (3,-1)

)**Identify the equation of the hyperbola with its center at (3, -1), its vertex at (5, -1), and an asymptote of 2y = 3x – 11.**

A) 9(x - 3) ^{2} - 4(y + 1) ^{2} -36 = 0 This looks promising :)

B) 4(y - 2) ^{2} - 9(y + 1) ^{2} = 36 There are no x'es so this one is no good.

^{2} + (x -1)^{2} + 36 = 0 This one has no solutions - so that is not it! (Those things can't add to 0)

D) 4(y +1)^{2} - 9(x -3)^{2} = 36 It is not this one

https://www.desmos.com/calculator/v4aixqae3g

**okay so it is A**

-------------------------------------------------------------------------------

Now I want to examine it

A) 9(x - 3) ^{2} - 4(y + 1) ^{2} -36 = 0 This looks promising :)

But I am much less sure about how to determine the vertex or the asymptote.

I saw Gino's post come up - he has probably explained it .

Melody
Nov 11, 2014

#1**+5 **

1) Since the center is at (3, -1) and a vertex is at (5, -1), it opens horizontally.

The general equation will be: (x - h)²/a² - (y - k)²/b² = 1

with center = (h, k), a = length of transverse radius, b = length of conjugate radius

For this problem: h = 3, k = -1, a = 2

The asymptote is 2y = 3x - 11 ---> y = (3/2)x - (11/2) ---> m = 3/2

Since m = 3/2, for every run of 2, it has a rise of 3 ---> Since a = 3, b = 2.

---> (x - 3)²/2² - (y - -1)²/2² = 1

Finish this, and you can choose the answer.

2) The general equaiton will be: (x - h)²/a² - (y - k)²/b² = 1

with center (h, k), a = x-radius, b = y-radius c = focal length

Since foci are at (1, -1) and (1, 5), the center will be at (1, 2) ---> h = 1, k = 2, c = 3

---> (x - 1)²/a² - (y - 2)²/b² = 1

Since the point (4, 2) is on the graph ---> (4 - 1)²/a² - (2 - 2)²/b² = 1,

solve this equation to find the value of a.

To find the value of b, use the equation: b² = a² + c².

(Use that equation if the major axis is vertical; if the major axis is horizontal, use a² = b + c².)

Place those values into the general equation to get your answer.

geno3141
Nov 11, 2014

#2**+5 **

Best Answer

I can see straight of from the formula that the centre will be (3,-1)

**Identify the equation of the hyperbola with its center at (3, -1), its vertex at (5, -1), and an asymptote of 2y = 3x – 11.**

A) 9(x - 3) ^{2} - 4(y + 1) ^{2} -36 = 0 This looks promising :)

B) 4(y - 2) ^{2} - 9(y + 1) ^{2} = 36 There are no x'es so this one is no good.

^{2} + (x -1)^{2} + 36 = 0 This one has no solutions - so that is not it! (Those things can't add to 0)

D) 4(y +1)^{2} - 9(x -3)^{2} = 36 It is not this one

https://www.desmos.com/calculator/v4aixqae3g

**okay so it is A**

-------------------------------------------------------------------------------

Now I want to examine it

A) 9(x - 3) ^{2} - 4(y + 1) ^{2} -36 = 0 This looks promising :)

But I am much less sure about how to determine the vertex or the asymptote.

I saw Gino's post come up - he has probably explained it .

Melody
Nov 11, 2014