Question

Find the equation of hyperbola with foci (+4,0) and length of the latus rectum is 12.
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Answer 1

Explanation:

Hi friend here is the solution.

Answer 2

Answer:

$x^{2}/{4} -y^{2} /{12} =1$

Explanation:

foci (+4,0) = (ae ,0) where a is length of transverse axis and e is the eccentricity of the hyperbola

latus rectum = 12

$\frac{2b^{2} }{a}$ = 12    where b is the length of conjugate axis

b² = 6a

we know,

e =$\sqrt{1+\frac{b^{2} }{a^{2} } }$ =

$\frac{4}{a}$ =$\sqrt{ 1+\frac{6a}{a^{2} } }$                    (∵ ae= 4)

squaring both the sides,

$\frac{16}{a^{2} } =1+\frac{6a}{a^{2} }$

a²+6a-16=0

on solving,we get

a=-8 (not selected as it is negative which gives imaginary under square root)

and

a=2 (taken)

so now

b²=6a=6×2=12

a²=4