Question

The eccentricity the hyperbola whose centre is (6,2) one focus is (4,2) and of eccentricity 2 is

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $2\sqrt{3}$
• D. $3\sqrt{2}$

Answer 1

The correct option is A

$\sqrt{2}$

$Weeliminatetinx$

$=\frac{a}{2}(t+\frac{1}{t}),\frac{a}{2}(t-\frac{1}{t})$

$\text{On squaring both the sides in the equations x}$

$=\frac{a}{2}(t+\frac{1}{t}),y=\frac{a}{2}(t-\frac{1}{t}),weget$

$\frac{4x^2}{a^2}=t^2+\frac{1}{t^2}+2$

$\Rightarrow \frac{4x^2}{a^2}-2=t^2+\frac{1}{t^2}...\left(1\right)$

$Also,\frac{4y^2}{a^2}=t^2+\frac{1}{t^2}-2$

$\Rightarrow \frac{4x^2}{a^2}+2=t^2+\frac{1}{t^2}...\left(2\right)$

$From\left(1\right)and\left(2\right),weget$

$\frac{4x^2}{a^2}-\frac{4y^2}{a^2}=4$

$\Rightarrow \frac{x^2}{a^2}-\frac{y^2}{a^2}=1$

$\text{This is the standard equation of hyperbola,where}{a}^2=b^2$

$Eccentricityofthehyperbola$

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