Question

Find the equation of the hyperbola whose vertices are at (-1, -5) and (-1, 1) with a focus at (-1, -7)
I have to show all of my work. Please help!

Answer 1

It is given that, the co-ordinates of vertices of hyperbola are (-1, -5) and (-1, 1)

Since, center is the mid-point of the vertices.

∴ Co-ordinates of Centre of hyperbola are [(-1-1)/2, (-5+1/)2] = [-2/2, -4/2] = [-1, -2]

Now, shifting the origin at (-1, -2) without rotating the axes.

Let X and Y denotes the new co-ordinates axes.

∴ X = x + 1 and Y = y + 2

⇒ x = X - 1 and y = Y - 2 --------------> (1)

The co-ordinates of vertices w.r.t. new axes are:

[∵ By putting (x = -1, y = -5) and (x = -1, y = 1) in (1)]

(-1+1, -5+2) and (-1+1, 1+2)

i.e., (0, -3) and (0, 3) i.e.,(0, ±3)

The co-ordinates of the focus w.r.t. new axes are: (-1+1, -7+2) i.e., (0, -5).

As the vertices and foci are on X-axis.

∴ Let equation of the hyperbola be: X²/a² - Y²/b² = 1 ---------------------->(2)

As vertices foci of hyperbola (1) are (0, ±a) and (0, ±ae) respectively.

∴ a = 3 and ae = -5

⇒ 3(e) = -5 ⇒ e = -5/3

b² = a²(e² - 1) = 3²[(-5/3)² - 1]

b² =9[(25/9) - 1]

b² =9[(25 - 9)/9]

b² =[16]

Substituting a² =9 and b² = 16 in (2), we have

X²/9 - Y²/16 = 1-------------------->(3)

which is the required equation of hyperbola w.r.t. new axes.

where: X = x + 1 and Y = y + 2

∴ From (1) and (3), we have (x + 1)²/9 - (y + 2)²/16 = 1.

16(x + 1)² - 9(y + 2)² = 144.

16(x² + 1 + 2x) - 9(y² + 4y + 4) = 144.

16x² + 16 + 32x - 9y² - 36y - 36 = 144

16x² - 9y² + 32x  - 36y - 20 = 144

16x² - 9y² + 32x  - 36y  = 164.

Therefore the equation of the hyperbola is 16x² - 9y² + 32x  - 36y  = 164.

Comments

The equation of hyperbola is .

Answer 2

It is given that, the co-ordinates of vertices of hyperbola are (-1, -5) and (-1, 1)

Since, center is the mid-point of the vertices.

∴ Co-ordinates of Centre of hyperbola are [(-1-1)/2, (-5+1/)2] = [-2/2, -4/2] = [-1, -2]

Center (h,k) , a,b are major and minor axis then h = -1 , k = -2

The equation of the hyper bola = (x -h)² / a² - (y - k)² / b² = 1

foci of hyperbola  (h , k± ae)

one focus = (-1 , -7)

The distance between two vertices : 2a = √[(-1 - (-1))² + (1- (-5))²]

                                                         2a = √(0² + 6²)

                                                         2a = 6

                                                         a = 3

The focus is (h , k + ae) = (-1 , -7)

k + ae = -7

But k = -2 and a = 3

-2 + 3e = - 7

3e = -5

e = -5 / 3

But b^2 = a²(e² - 1)

Substitute a = 3 and e = -5 / 3

b² = 9((25 / 9) - 1)

b² = 25 - 9

b² = 16

 The equation of the hyperbola = (x -h)² / a² - (y - k)² / b² = 1

Substitute a² = 9 , b² = 16 , h = -1 and k = -2 in the above equation

(x +1)² / 9 - (y + 2)² / 16 = 1

Multiply each side by 144

16(x + 1)² - 9(y + 2)² = 144

16x² + 32x + 16 - 9y² -36y - 36 = 144

16x² - 9y² + 32x - 36y  = 164.

Therefore the equation of the hyperbola is 16x² - 9y² + 32x  - 36y  = 164.

Comments

The equation of hyperbola is .

Answer 3

The vertices of the hyperbola are (-1, -5) and (-1, 1) and its focus at  (-1, - 7) .

Since the x - coordinate is constant in the vertices and foci.

This is vertical hyperbola.

Standard form of vertical hyperbola is

"a " is the number in the denominator of the positive term

If the y -term is positive, then the hyperbola is vertical

a = semi-transverse axis , b = semi-conjugate axis

center: (h, k ) Vertices: (h , k + a), (h , k - a )

Foci: (h , k + c ), (h , k - c )

So the x coodinate of the center of hyperbola is  - 1.

vertices: (- 1, - 5) and (-1,1)

k + a = - 5 ----> (1)

k - a = 1 ------> (2)

Add the equations (1) & (2).

2k = - 4

k = - 2

So y coordinate of center is - 2.

(h , k ) = (-1, -2)

Substitute the k  value in (1).

- 2 + a = - 5

a = - 5 + 2

a  = - 3

Since a  is positive term.

a  = 3

Foci: (h , k + c ), (h , k - c )

(-1, - 2 + c), (-1, -2 - c)

- 2 + c = -7

c = - 7 + 2

c = - 5

So c = - 5

c² = a² + b²

(5)² = (3)² + b²

25 - 9 = b²

b = ± 4

Therfore b = 4

Substitute (h ,k ) ,a  and b  values in standard form of vertical parabola.

Therefore the equation of hyperbola is