The equation of the normal at the point (6, 4 | vedclass

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(A) The given hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 3$. Dividing by $3$,we get $\frac{x^2}{27} - \frac{y^2}{48} = 1$.Here,$a^2 = 27$ and $b^2

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$3x + 8y = 50$

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(A) The given hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 3$. Dividing by $3$,we get $\frac{x^2}{27} - \frac{y^2}{48} = 1$.Here,$a^2 = 27$ and $b^2 = 48$.The equation of the normal at point $(x_1, y_1)$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\frac{a^2(x - x_1)}{x_1} + \frac{b^2(y - y_1)}{y_1} = 0$.Substituting $(x_1, y_1) = (6, 4)$,$a^2 = 27$,and $b^2 = 48$:$\frac{27(x - 6)}{6} + \frac{48(y - 4)}{4} = 0$$\frac{9(x - 6)}{2} + 12(y - 4) = 0$Multiplying by $2$:$9(x - 6) + 24(y - 4) = 0$$9x - 54 + 24y - 96 = 0$$9x + 24y = 150$Dividing by $3$:$3x + 8y = 50$.

The equation of the normal at the point $(6, 4)$ on the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 3$ is:

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