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# Identifying Equations

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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
+92254
+5

)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.

C) (y + 3)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 :)

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

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
Sort:

#1
+17721
+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.)

geno3141  Nov 11, 2014
#2
+92254
+5

)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.

C) (y + 3)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 :)

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

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

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