Question

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

विकल्प

• 3 (x − 6)² − (y −2)² = 3

• (x − 6)² − 3 (y − 2)² = 1

• (x − 6)² − 2 (y −2)² = 1

• 2 (x − 6)² − (y − 2)² = 1

Answer 1

3 (x − 6)² − (y −2)² = 3

The equation of the hyperbola with centre (x₀,y₀) is given by $$\frac{{(x-x_0)}^2}{a^2}-\frac{{(y-y_0)}^2}{b^2}=1$$

Focus = $$(ae+x_0,y_0)$$

$$\therefore ae=-2$$

$$\Rightarrow a=-1$$

$$b^2={\left(2\right)}^2-a^2$$

$$\Rightarrow b^2={(-2)}^2-{(-1)}^2$$

$$\Rightarrow b^2=3$$

$$\therefore \frac{{(x-6)}^2}{1}-\frac{{(y-2)}^2}{3}=1$$

$$\Rightarrow 3{(x-6)}^2-{(y-2)}^2=3$$