Correct Answer - Option 3 :
\((±\sqrt{20} , 0)\)
Concept:
Hyperbola:
Equation
|
\(\rm \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
|
\(\rm\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\)
|
Equation of Transverse axis
|
y = 0
|
x = 0
|
Equation of Conjugate axis
|
x = 0
|
y = 0
|
Length of Transverse axis
|
2a
|
2a
|
Length of Conjugate axis
|
2b
|
2b
|
Vertices
|
(± a, 0)
|
(0, ± a)
|
Focus
|
(± ae, 0)
|
(0, ± ae)
|
Directrix
|
x = ± a/e
|
y = ± a/e
|
Centre
|
(0, 0)
|
(0, 0)
|
Eccentricity
|
\(\rm \sqrt{1+\frac{b^{2}}{a^{2}}}\)
|
\(\rm \sqrt{1+\frac{b^{2}}{a^{2}}}\)
|
Length of Latus rectum
|
\(\rm \frac{2b^{2}}{a}\)
|
\(\rm \frac{2b^{2}}{a}\)
|
Focal distance of the point (x, y)
|
ex ± a
|
ey ± a
|
Calculation:
Given: \(\rm\frac{x^{2}}{16}-\frac{y^{2}}{4}=1\)
It can be written as,
\(\rm⇒ \frac{x^{2}}{16}-\frac{y^{2}}{4}=\frac{x^{2}}{4^{2}}-\frac{y^{2}}{2^{2}}=1\)
On comparing with the above equation with the general equation of the hyperbola, \(\rm \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
⇒ a = 4 and b = 2
As we know that eccentricity of hyperbola is given by:\(\rm \sqrt{1+\frac{b^{2}}{a^{2}}}\).
\(\rm ⇒ \sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{2^{2}}{4^{2}}}=\frac{\sqrt{20}}{4}\)
As we know that, the coordinates of the foci of the hyperbola \(\rm \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is (± ae, 0).
\(\rm ⇒(± ae, 0)=(± 4\times\frac{\sqrt{20}}{4}, 0)=(±\sqrt{20} , 0)\)
Hence, the correct option is 3.