If S equiv frac{x^2}{k-7}+frac{y^2}{11-k}-1=0 | vedclass

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(D) Given $S = \frac{x^2}{k-7} + \frac{y^2}{11-k} = 1$. For $k=9$: $\frac{x^2}{2} + \frac{y^2}{2} = 1$,which is a circle with radius $\sqrt{2}$. State

Vedclass · Accepted Answer

$S=0$ represents a hyperbola with eccentricity $\sqrt{\frac{3}{2}}$ when $k=13$

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(D) Given $S = \frac{x^2}{k-7} + \frac{y^2}{11-k} = 1$. For $k=9$: $\frac{x^2}{2} + \frac{y^2}{2} = 1$,which is a circle with radius $\sqrt{2}$. Statement $A$ is correct. For $k=10$: $\frac{x^2}{3} + \frac{y^2}{1} = 1$,which is an ellipse with $a^2=3, b^2=1$. Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}}$. Statement $B$ is correct. For $k=12$: $\frac{x^2}{5} - \frac{y^2}{1} = 1$,which is a hyperbola with $a^2=5, b^2=1$. Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{5}} = \sqrt{\frac{6}{5}}$. Statement $C$ is correct. For $k=13$: $\frac{x^2}{6} - \frac{y^2}{2} = 1$,which is a hyperbola with $a^2=6, b^2=2$. Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{2}{6}} = \sqrt{\frac{8}{6}} = \sqrt{\frac{4}{3}}$. Statement $D$ is incorrect.

If $S \equiv \frac{x^2}{k-7}+\frac{y^2}{11-k}-1=0, k \in R-\{7,11\}$,then which one of the following statements is incorrect?

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