**Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola (x**^{2})/_{16}– (y^{2})/_{9}= 1

Solution:

The given equation is (x^{2})/_{16} – (y^{2})/_{9} = 1

or

[(x^{2})/(4^{2})] – [(y^{2})/(3^{2})] = 1

On comparing this equation with the standard equation of hyperbola [(x^{2})/(a^{2})] – [(y^{2})/(b^{2})] = 1 i.e., , we obtain a = 4 and b = 3.

We know that a^{2} + b^{2} = c^{2}

c^{2} = 4^{2} + 3^{2} = 16+ 9 = 25

c = 5

Therefore,

The coordinates of the foci are (±5, 0).

The coordinates of the vertices are (±4, 0).

Eccentricity, e = ^{c}/_{a} = ^{5}/_{4}

Length of latus rectum = [2b^{2}]/_{a} = ^{2×9}/_{4} = ^{9}/_{2}

**Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola (y**^{2})/_{9}– (x^{2})/_{27}= 1

Solution

The given equation is (y^{2})/_{9} – (x^{2})/_{27} = 1 or (y^{2})/(3^{2}) – (x^{2})/(√27)^{2} = 1

On comparing this equation with the standard equation of hyperbola [(x^{2})/(a^{2})] – [(y^{2})/(b^{2})] = 1 we obtain a = 3 and b = √27

We know that a^{2} + b^{2} = c^{2}

c^{2} = 3^{2} + (√27)^{2} = 9 + 27 = 36

c = 6

Therefore,

The coordinates of the foci are (0, ±6).

The coordinates of the vertices are (0, ±3).

Eccentricity e = ^{c}/_{a} = ^{6}/_{3} = 2

Length of latus rectum = (2b^{2})/_{a} = ^{2×27}/_{3} = 18

**Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9y**^{2}– 4x^{2}= 36

Solution:

The given equation is 9y^{2} – 4x^{2} = 36.

It can be written as 9y^{2} – 4x^{2} = 36

or

(y^{2})/_{4} – (x^{2})/_{9} = 1

(y^{2})/(2^{2}) – (x^{2})/(3^{2}) = 1 ——–(1)

On comparing equation (1) with the standard equation of hyperbola [(x^{2})/(a^{2})] – [(y^{2})/(b^{2})] = 1

we obtain a = 2 and b = 3.

We know that a^{2} + b^{2} = c^{2}

c^{2} = 4 + 9 = 13

c = √13

Therefore,

The coordinates of the foci are (0, ±√13)

The coordinates of the vertices are (0, ±2)

Eccentricity, e = ^{c}/_{a} = ^{√13}/_{2}

Length of latus rectum = (2b^{2})/_{a} = ^{2×9}/_{2} = 9

**Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x**^{2}– 9y^{2}= 576

Solution:

The given equation is 16x^{2} – 9y^{2} = 576.

It can be written as

16x^{2} – 9y^{2} = 576

[(x^{2})/36] – [(y^{2})/64] = 1

[(x^{2})/6^{2}] – [(y^{2})/8^{2}] = 1 ————(1)

On comparing equation (1) with the standard equation of hyperbola i.e., [(x^{2})/a^{2}] – [(y^{2})/b^{2}] = 1 , we obtain a = 6 and b = 8.

We know that a^{2} + b^{2} = c^{2}.

c^{2} = 36 + 64 = 100

c = 10

Therefore,

The coordinates of the foci are (±10, 0).

The coordinates of the vertices are (±6, 0).

Eccentricity, e = ^{c}/_{a} = ^{10}/_{6} = ^{5}/_{3}

Length of latus rectum = (2b^{2})/_{a} = ^{2×64}/_{6} = ^{64}/_{3}

**Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y**^{2}– 9x^{2}= 36

solution:

The given equation is 5y^{2} – 9x^{2} = 36.

(y^{2})/_{[36/5]} – (x^{2})/_{4} = 1

(y^{2})/_{[6/√5]}^{2 } – (x^{2})/(2^{2}) = 1 —————(1)

On comparing equation (1) with the standard equation of hyperbola [(y^{2})/(a^{2})] – [(x^{2})/(b^{2})] = 1, we obtain a =6/√5 and b = 2.

We know that a^{2} + b^{2} = c^{2}

c^{2} = ^{36}/_{5} + 4 = ^{56}/_{5}

c = √(^{56}/_{5}) = ^{2}^{√14}/_{√5}

Therefore, the coordinates of the foci are (0, ±^{2}^{√14}/_{√5})

The coordinates of the vertices are (0, ±^{6}/_{√5})

Eccentricity, e = ^{c}/_{a} = [^{2}^{√14}/_{√5}]/[^{ 6}/_{√5}] = ^{√14}/_{3}

Length of latus rectum = 2b^{2}/_{a} = ^{2×4}/_{(6/√5)} = ^{4x√5}/_{3}

**Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49y**^{2}– 16x^{2}= 784

solution:

The given equation is 49y^{2} – 16x^{2} = 784.

It can be written as

49y^{2} – 16x^{2} = 784

(y^{2})/_{16} – (x^{2})/_{49} = 1

(y^{2})/(4^{2}) – (x^{2})/(7^{2}) = 1 ————(1)

On comparing equation (1) with the standard equation of hyperbola (y^{2})/(a^{2}) – (x^{2})/(b^{2}) = 1 i.e., , we obtain a = 4 and b = 7.

We know that a^{2} + b^{2} = c^{2}

c^{2} = 16 + 49 = 65

c = √65

Therefore,

The coordinates of the foci are (0, ±√65)

The coordinates of the vertices are (0, ±4).

Eccentricity, e = ^{c}/_{a} = ^{√65}/_{4}

Length of latus rectum = (2b^{2})/_{a} = ^{2×49}/_{4} = ^{49}/_{2}

**Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci (±3, 0)**

Solution:

Vertices (±2, 0), foci (±3, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form (x^{2})/(a^{2}) – (y^{2})/(b^{2}) = 1.

Since the vertices are (±2, 0), a = 2.

Since the foci are (±3, 0), c = 3.

We know that a^{2} + b^{2} = c^{2}

2^{2} + b^{2 }= 3^{2}

b^{2} = 9 – 4 = 5

Thus, the equation of the hyperbola is (x^{2})/(4) – (y^{2})/(5) = 1

**Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±5), foci (0, ±8)**

Solution:

Vertices (0, ±5), foci (0, ±8)

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form (y^{2})/(a^{2}) – (x^{2})/(b^{2}) = 1.

Since the vertices are (0, ±5), a = 5.

Since the foci are (0, ±8), c = 8.

We know that a^{2} + b^{2} = c^{2}

5^{2} + b^{2} = 8^{2}

b^{2} = 64 – 25 = 39

Thus, the equation of the hyperbola is .

**Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±3), foci (0, ±5)**

Solution:

Vertices (0, ±3), foci (0, ±5)

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form (y^{2})/(a^{2}) – (x^{2})/(b^{2}) = 1.

Since the vertices are (0, ±3), a = 3.

Since the foci are (0, ±5), c = 5.

We know that a^{2} + b^{2} = c^{2}

∴3^{2} + b^{2} = 5^{2}

⇒ b^{2} = 25 – 9 = 16

Thus, the equation of the hyperbola is (y^{2})/(9) – (x^{2})/(16) = 1

**Find the equation of the hyperbola satisfying the give conditions: Foci (±5, 0), the transverse axis is of length 8.**

Solution:

Foci (±5, 0), the transverse axis is of length 8.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form (x^{2})/(a^{2}) – (y^{2})/(b^{2}) = 1.

Since the foci are (±5, 0), c = 5.

Since the length of the transverse axis is 8, 2a = 8

⇒ a = 4.

We know that a^{2} + b^{2} = c^{2}.

∴4^{2} + b^{2} = 5^{2}

⇒ b² = 25 – 16 = 9

b = 3

Thus, the equation of the hyperbola is .(x^{2})/16 – (y^{2})/9 = 1

**Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±13), the conjugate axis is of length 24.**

Solution:

Foci (0, ±13), the conjugate axis is of length 24.

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form (y^{2})/(a^{2}) – (x^{2})/(b^{2}) = 1.

Since the foci are (0, ±13), c = 13.

Since the length of the conjugate axis is 24, 2b = 24 ⇒ b = 12.

We know that a^{2} + b^{2} = c^{2}

∴a^{2} + 12^{2} = 13^{2}

⇒ a^{2} = 169 – 144 = 25

Thus, the equation of the hyperbola is (y^{2})/_{25} – (x^{2})/_{144} = 1

**Find the equation of the hyperbola satisfying the give conditions: Foci (±3√5, 0) , the latus rectum is of length 8.**

Solution:

Foci (±3√5, 0) the latus rectum is of length 8.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form (x^{2})/(a^{2}) – (y^{2})/(b^{2}) = 1

Since the foci are

(±3√5, 0)c = ±3√5

Length of latus rectum = 8

2b^{2}/_{a} = 8

b^{2} = 4a

We know that, a^{2} + b^{2} = c^{2}

a^{2} + 4a = 45

a^{2} + 4a – 45 = 0

a^{2} + 9a – 5a – 45 = 0

(a + 9)(a – 5) = 0

a = -9 or 5

Since a is non- negative, a = 5

b^{2} = 4a = 4 x 5 = 20

Thus, the equation of the hyperbola is x^{2}/_{25} – y^{2}/_{20} = 1

**Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus rectum is of length 12**

Solution:

Foci (±4, 0), the latus rectum is of length 12.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form x^{2}/a^{2} – y^{2}/b^{2} = 1.

Since the foci are (±4, 0), c = 4.

Length of latus rectum = 12

2b^{2}/_{a}= 12

b^{2} = 6a

We know that a^{2} + b^{2} = c^{2}

∴a^{2} + 6a = 16

a^{2} + 8a – 6a – 16 = 0

(a + 8)(a – 2) = 0

a = -8 or 2

Since a is non negative, a = 2

b^{2} = 6a = 6 x 2 = 12

Thus, the equation of the hyperbola is x^{2}/_{4} – y^{2}/_{12} = 1

**Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0), e =**^{4}/_{3}

Solution:

Vertices (±7, 0), e = ^{4}/_{3}

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form x^{2}/a^{2} – y^{2}/b^{2} = 1

Since the vertices are (±7, 0), a = 7.

It is given that e = ^{4}/_{3} , we know e = ^{c}/_{a}

^{c}/_{a} = ^{4}/_{3}

^{c}/_{7} = ^{4}/_{3}

c = ^{28}/_{3}

We know that a^{2} + b^{2} = c^{2}.

7^{2} + b^{2} = (^{28}/_{3})^{2}

b^{2} = ^{784}/_{9} – 49

b^{2} = ^{(784 – 441)}/_{9} = ^{343}/_{9}

Thus the equation of the hyperbola x^{2}/_{49} – (9y^{2})/_{343} = 1

**Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±√10) , passing through (2, 3)**

Solution:

Foci (0, ±√10) , passing through (2, 3)

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form y^{2}/a^{2} – x^{2}/b^{2} = 1

Since the foci are (0, ±√10)

c = √10

We know that a^{2} + b^{2} = c^{2}

∴ a^{2} + b^{2} = 10

⇒ b^{2} = 10 – a^{2} ————- (1)

Since the hyperbola passes through point (2, 3)

^{9}/a^{2} – ^{4}/b^{2} = 1 —————-(2)

From equations (1) and (2), we obtain

^{9}/a^{2} – ^{4}/(10-a^{2}) = 1

9(10 – a^{2}) – 4a^{2} = a^{2}(10 – a^{2})

90 – 9a^{2} – 4a^{2} = a^{2}10 – a^{4}

a^{4}– 23a^{2} + 90 = 0

a^{4}– 18a^{2}– 5a^{2} + 90 = 0

a^{2}(a^{2} – 18)-5(a^{2} – 18) = 0

a^{2} = 18 or 5

In hyperbola c > a i.e., c^{2} > a^{2}

a^{2} = 5

b^{2} = 10 – a^{2} = 10 – 5 = 5

Thus the equation of the hyperbola is y^{2}/_{5} – x^{2}/_{5} = 1