Question

Find the hyperbola satisfying the following conditions: 

1. Vertices (+2, 0), foci (±3, 0)

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

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

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

5. Vertices (±7, 0) , e = \(\frac{4}{3}\).

Answer 1

1. Since Vertices are (±2, 0) the standard form of hyperbola is

\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.\)

⇒ a = 2

Given; foci (±3, 0) 

⇒ c = 3

⇒ c² = a² + b² 

⇒ 9 = 4 + b² ⇒ b² = 5

The equation of the hyperbola;

\(\frac{x^2}{4} - \frac{y^2}{5} = 1\)

2. Since Foci are (±5,0) the standard form of hyperbola is

\(\frac{x^2}{a^2} - \frac{y^2}{b^2}= 1\)

⇒ c = 5

Given; the transverse axis is of length 8.

⇒ 2a = 8 ⇒ a = 4

⇒ c² = a² + b² ⇒ 25 = 16 + b² ⇒ b² = 9

The equation of the hyperbola;

\(\frac{x^2}{19} - \frac{y^2}{9}\) = 1

3. Since Foci are(0, ±13)the standard form of hyperbola is

\(\frac{y^2}{a^2} - \frac{x^2}{b^2}= 1\)

⇒ c = 13

Given; the conjugate axis is of length 24.

⇒ 2b = 24 ⇒ b = 12

⇒ c² = a² + b² ⇒ 169 = a² + 144 ⇒ a² = 25.

The equation of the hyperbola;

\(\frac{y^2}{25} - \frac{x^2}{12}= 1\)

4. Since Foci are(±3 √5, 0) the standard form of hyperbola is

⇒ c = 3√5

Given; the latus rectum is of length 8.

⇒ \(\frac{2b^2}{a} \) = 8 

⇒ b² = 4a

⇒ c² = a² + b²

⇒ 45 = a² + 4a 

⇒ a² + 4a – 45 = 0

⇒ (a + 9)(a – 5) = 0 

⇒ a = -9, 5

a = -9 is not possible

⇒ a = 5 ⇒ b² = 20

The equation of the hyperbola;

\(\frac{x^2}{25}-\frac{y^2}{20} = 1\)

5. Since Vertices are (±7, 0) the standard form of hyperbola is

\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. \)

⇒ a = 7

Given;

e = \(\frac{4}{3} \Rightarrow \frac{c}{a} = \frac{4}{3}\)

\(\Rightarrow \frac{c}{7} = \frac{4}{3} \Rightarrow c = \frac{28}{3}\)

⇒ c² = a² + b²

\(\Rightarrow \big(\frac{28}{3}\big)^2\) = 49 + b² 

⇒ b² = \(\frac{343}{9}\)

The equation of the hyperbola;

⇒ \(\frac{x^2}{49} - \frac{9y^2}{343} = 1\)