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If $S \equiv \frac{\mathrm{X}^2}{\mathrm{k}-7}+\frac{\mathrm{y}^2}{11-\mathrm{k}}-1=0, \mathrm{k} \in \mathrm{R}-\{7,11\}$, then which one of the following statements is incorrect?
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The correct answer is:
$S=0$ represents a hyperbola with eccentricity $\sqrt{\frac{6}{5}}$ when $\mathrm{k}=12$
Given that $S=\frac{x^2}{k-7}+\frac{y^2}{11-k}-1=0...(i)$
Put $k=13$
$\Rightarrow \frac{x^2}{6}-\frac{y^2}{2}-1=0 \Rightarrow \frac{x^2}{6}-\frac{y^2}{2}=1$
Represent equation of hyperbola, where $a^2=6$ and $b^2=2$.
Now, $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{2}{6}}=\frac{2}{\sqrt{3}}$
Put $k=12$, we get $\frac{x^2}{5}-\frac{y^2}{1}=1$
Represent equation of hyperbola where $a^2=5$ and $b^2=1$.
Now, $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{5}}=\sqrt{\frac{6}{5}}$.
Put $k=13$
$\Rightarrow \frac{x^2}{6}-\frac{y^2}{2}-1=0 \Rightarrow \frac{x^2}{6}-\frac{y^2}{2}=1$
Represent equation of hyperbola, where $a^2=6$ and $b^2=2$.
Now, $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{2}{6}}=\frac{2}{\sqrt{3}}$
Put $k=12$, we get $\frac{x^2}{5}-\frac{y^2}{1}=1$
Represent equation of hyperbola where $a^2=5$ and $b^2=1$.
Now, $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{5}}=\sqrt{\frac{6}{5}}$.
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