Eccentricity of hyperbola x2-k-y2-k-1k-0 is -A- -1-1-k2B- -1-kC- -1-kD- -1-1-k2

The correct option is D Given equation can be rewritten as y^2/k^2 x^2/( k) = 1 ( k gt;0 ) e^2 = 1+ ( k)/k^2 = 1 1/k⇒ e =√(1 1/k) ...

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Byju's Answer

Standard XII

Mathematics

Point Form of Tangent: Hyperbola

Eccentricity ...

Question

Eccentricity of hyperbola $\frac{x^{2}}{k} + \frac{y^{2}}{k} = 1 (k < 0)$ is :

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Solution

The correct option is D
Given equation can be rewritten as $\frac{y^2}{k^2}- \frac{x^2}{(-k)}$ $= 1 (- k > 0)$

$e^{2} = 1 + \frac{(- k)}{k^{2}}$ $= 1 - \frac{1}{k} \Rightarrow e = \sqrt{1 - \frac{1}{k}}$

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Eccentricity of hyperbola x2k+y2k=1(k<0) is :

The correct option is D Given equation can be rewritten as \(\frac{y^2}{k^2}- \frac{x^2}{(-k)}\) =1(−k>0) e2=1+(−k)k2 =1−1k⇒e=√1−1k

Eccentricity of hyperbola $\frac{x^{2}}{k} + \frac{y^{2}}{k} = 1 (k < 0)$ is :

The correct option is D
Given equation can be rewritten as \(\frac{y^2}{k^2}- \frac{x^2}{(-k)}\) $= 1 (- k > 0)$

$e^{2} = 1 + \frac{(- k)}{k^{2}}$ $= 1 - \frac{1}{k} \Rightarrow e = \sqrt{1 - \frac{1}{k}}$