If the line y = mx + 7sqrt{3} is normal to th | vedclass

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(A) The equation of the hyperbola is $\frac{x^2}{24} - \frac{y^2}{18} = 1$,where $a^2 = 24$ and $b^2 = 18$.The condition for the line $y = mx + c$ to

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$\frac{2}{\sqrt{5}}$

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(A) The equation of the hyperbola is $\frac{x^2}{24} - \frac{y^2}{18} = 1$,where $a^2 = 24$ and $b^2 = 18$.The condition for the line $y = mx + c$ to be a normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = \frac{m^2(a^2 + b^2)^2}{a^2m^2 - b^2}$.Here,$c = 7\sqrt{3}$,so $c^2 = 49 	imes 3 = 147$.Substituting the values: $147 = \frac{m^2(24 + 18)^2}{24m^2 - 18} = \frac{m^2(42)^2}{24m^2 - 18} = \frac{1764m^2}{24m^2 - 18}$.Dividing by $147$: $1 = \frac{12m^2}{24m^2 - 18}$.$24m^2 - 18 = 12m^2$ $\Rightarrow 12m^2 = 18$ $\Rightarrow m^2 = \frac{18}{12} = \frac{3}{2}$.Wait,re-evaluating the normal condition: The normal at $(x_0, y_0)$ is $\frac{a^2x}{x_0} + \frac{b^2y}{y_0} = a^2 + b^2$. Comparing with $y = mx + c$,we get $m = -\frac{a^2y_0}{b^2x_0}$.Using the slope form $m = \pm \frac{c}{\sqrt{a^2 - b^2m^2}}$ is not applicable here. The correct condition for $y = mx + c$ to be normal is $c^2 = \frac{m^2(a^2 + b^2)^2}{a^2m^2 - b^2}$ is incorrect. The correct condition is $c^2 = \frac{m^2(a^2 + b^2)^2}{a^2m^2 - b^2}$ is for ellipse. For hyperbola,it is $c^2 = \frac{m^2(a^2 + b^2)^2}{a^2m^2 - b^2}$ is wrong. The correct condition is $c = \pm \frac{m(a^2 + b^2)}{\sqrt{a^2 - b^2m^2}}$.$c^2(a^2 - b^2m^2) = m^2(a^2 + b^2)^2$ $\Rightarrow 147(24 - 18m^2) = m^2(42)^2$ $\Rightarrow 147(24 - 18m^2) = 1764m^2$.Divide by $147$: $24 - 18m^2 = 12m^2$ $\Rightarrow 30m^2 = 24$ $\Rightarrow m^2 = \frac{24}{30} = \frac{4}{5}$.Thus,$m = \pm \frac{2}{\sqrt{5}}$.

If the line $y = mx + 7\sqrt{3}$ is normal to the hyperbola $\frac{x^2}{24} - \frac{y^2}{18} = 1$,then a value of $m$ is

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