$\frac{x^{2}}{r^{2} - r - 6} + \frac{y^{2}}{r^{2} - 6 r + 5} = 1$ will represents the ellipse, if r lies in the interval:

(- \infty, - 2)

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(3, \infty)

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(5, \infty)

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(1, \infty)

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Solution

The correct options are
A $(- \infty, - 2)$
D $(5, \infty)$
$r^{2} - r - 6 > 0$ and $r^{2} - 6 r - 5 > 0$
$\Rightarrow (r - 3) (r + 2) > 0$ and $(r - 1) (r - 5) > 0$
$\Rightarrow (r < - 2$ or $r > 3)$ and $(r < 1$ or $r > 5)$
$\Rightarrow r < - 2$ or $r > 5$
Also $r^{2} - r - 6 \neq r^{2} - 6 r + 5$
$\Rightarrow r \neq \frac{11}{5}$

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Similar questions

\frac{x^{2}}{2 - r} + \frac{y^{2}}{r - 5} + 1 = 0

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\frac{x^{2}}{2 - r} + \frac{y^{2}}{r - 5} + 1 = 0

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