A-2-b-2-z-2-8x-4y - a-2 - -x- Then x equalsb-2-z-2-8x-4y-b-2-z-2-8x-4y-b-2-z-2-8x-4y-b-2-z-2-8x-4y-

The correct option is D b^2 z^2 8x+4y a^2 b^2+z^2+8x 4y = a^2 ( b^2 z^2 8x+4y ) ∴ x= b^2 z^2 8x+4y ...

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Addition of Brackets

a^2-b^2+z^2+8...

Question

$a^{2} - b^{2} + z^{2} + 8 x - 4 y$ = $a^{2} - (x)$ . Then $x$ equals

$b^{2} + z^{2} - 8 x - 4 y$

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$b^{2} + z^{2} - 8 x + 4 y$

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$b^{2} - z^{2} - 8 x + 4 y$

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$- b^{2} + z^{2} + 8 x - 4 y$

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$a^{2} - b^{2} + z^{2} + 8 x - 4 y$ = $a^{2} - (x)$ . Then $x$ equals

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1, - \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

\in R^{+}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

- \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

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