Correct Answer – Option 2 : \(\rm {(x-1)^2\over 4}-{y^2\over 12} = 1\)

Concept:

The standard equation of a hyperbola:

\(\rm {(x-h)^2\over a^2}-{(y-k)^2\over b^2} = 1\)

where 2a and 2b are the length of the transverse axis and conjugate axis respectively and centre (h, k)

Note: The centre is the midpoint of the 2 foci.

The eccentricity = \(\rm \sqrt{a^2+b^2}\over a\)

Length of latus recta = \(\rm 2b^2 \over a\)

Distance from center to focus = \(\rm \sqrt{a^2+b^2}\)

Calculation:

Given foci are (5,0) and (-3,0)

Center = \(\rm \left({5+(-3)\over2},{0+0\over2}\right)\) = (1, 0)

Now distance of focus from the center = \(\rm \sqrt{a^2+b^2}\)

⇒ \(\rm \sqrt{(5-1)^2+(0-0)^2} = \rm \sqrt{a^2+b^2}\)

⇒ a² + b² = 16               …(i)

The eccentricity = \(\rm \sqrt{a^2+b^2}\over a\)

⇒ 2 = \(\rm \sqrt{16}\over a\) 

⇒ a = 2

Putting it in (i)

⇒ 4 + b² = 16

⇒ b² = 12

a² = 4

The equation of the hyperbola

\(\rm {(x-h)^2\over a^2}-{(y-k)^2\over b^2} = 1\)

⇒ \(\rm {(x-1)^2\over 4}-{(y-0)^2\over 12} = 1\)

⇒ \(\boldsymbol{\rm {(x-1)^2\over 4}-{y^2\over 12} = 1}\)