Question

Question
find the equation of hyperbola whose
focii are (+ 4,0) and the length of latus
rectum is 12 ?​

Answer 1

Answer:

Please see the attachment

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the:Question}}}}$

Foci of hyperbola are form (±a , 0)

Hence,

It is a horizontal ellipse

Now,

Assume

${\boxed{\sf\:{Equation=\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1}}}$

Assume,

Foci = (±c , 0)

Then,

c = 4

Therefore,

Foci = (±4 , 0)

Now,

c = 4

c² = 16

a² + b² = 16 ..... (1)

Length of Latus rectum = 12

$\tt{\rightarrow\dfrac{2b^2}{a}=12}$

b² = 6a ...... (2)

From (1) and (2)

a² + 6a = 16

a² + 6a - 16 = 0

(a + 8)(a - 2) = 0

a = 2

${\boxed{\sf\:{Substitute\;value\;in\;(2)}}}$

b² = (6 × 2)

= 12

Hence,

a² = 2² = 4

b² = 12

Hence,

$\Large{\boxed{\sf\:{Equation=\dfrac{x^2}{4}-\dfrac{y^2}{12}=1}}}$