If a hyperbola passes through the point P(10, | vedclass

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(C) The standard equation of a hyperbola with vertices at $(\pm a, 0)$ is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Given $a = 6$,the equation is $\frac

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$2x + 5y = 100$

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(C) The standard equation of a hyperbola with vertices at $(\pm a, 0)$ is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Given $a = 6$,the equation is $\frac{x^2}{36} - \frac{y^2}{b^2} = 1$.Since the hyperbola passes through $P(10, 16)$,we have $\frac{100}{36} - \frac{256}{b^2} = 1$.$\frac{25}{9} - 1 = \frac{256}{b^2} \implies \frac{16}{9} = \frac{256}{b^2} \implies b^2 = \frac{256 	imes 9}{16} = 144$.The equation of the hyperbola is $\frac{x^2}{36} - \frac{y^2}{144} = 1$.The equation of the normal at $(x_1, y_1)$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$.Substituting $a^2 = 36, b^2 = 144, x_1 = 10, y_1 = 16$:$\frac{36x}{10} + \frac{144y}{16} = 36 + 144 = 180$.$3.6x + 9y = 180$.Dividing by $1.8$,we get $2x + 5y = 100$.

If a hyperbola passes through the point $P(10, 16)$ and it has vertices at $(\pm 6, 0)$,then the equation of the normal to it at $P$ is

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