The eccentricity of the hyperbola $\frac{{\sqrt {1999} }}{3}({x^2} - {y^2}) = 1$ is
The eccentricity of the hyperbola $\frac{{\sqrt {1999} }}{3}({x^2} - {y^2}) = 1$ is
- A
$\sqrt 3 $
- B
$\sqrt 2 $
- C
$2$
- D
$2\sqrt 2 $
Similar Questions
Let the tangent to the parabola $y^2=12 x$ at the point $(3, \alpha)$ be perpendicular to the line $2 x+2 y=3$.Then the square of distance of the point $(6,-4)$from the normal to the hyperbola $\alpha^2 x^2-9 y^2=9 \alpha^2$at its point $(\alpha-1, \alpha+2)$ is equal to $........$.
Let the tangent to the parabola $y^2=12 x$ at the point $(3, \alpha)$ be perpendicular to the line $2 x+2 y=3$.Then the square of distance of the point $(6,-4)$from the normal to the hyperbola $\alpha^2 x^2-9 y^2=9 \alpha^2$at its point $(\alpha-1, \alpha+2)$ is equal to $........$.
- [JEE MAIN 2023]
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