The normal to the hyperbola frac{x^{2}}{a^{2} | vedclass

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Question

(C) Given the hyperbola equation $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$. Since the point $(8, 3\sqrt{3})$ lies on the hyperbola,we substitute the coo

Vedclass · Accepted Answer

$(-1, 9\sqrt{3})$

Vedclass · Answer

(C) Given the hyperbola equation $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$. Since the point $(8, 3\sqrt{3})$ lies on the hyperbola,we substitute the coordinates into the equation: $\frac{64}{a^{2}}-\frac{(3\sqrt{3})^{2}}{9}=1$ $\frac{64}{a^{2}}-\frac{27}{9}=1$ $\frac{64}{a^{2}}-3=1$ $\Rightarrow \frac{64}{a^{2}}=4$ $\Rightarrow a^{2}=16$. The equation of the normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at point $(x_{1}, y_{1})$ is given by $\frac{a^{2}x}{x_{1}}+\frac{b^{2}y}{y_{1}}=a^{2}+b^{2}$. Substituting $a^{2}=16, b^{2}=9, x_{1}=8, y_{1}=3\sqrt{3}$: $\frac{16x}{8}+\frac{9y}{3\sqrt{3}}=16+9$ $2x+\sqrt{3}y=25$. Checking the options: For option $A$: $2(15)+\sqrt{3}(-2\sqrt{3}) = 30-6 = 24 
eq 25$. For option $B$: $2(9)+\sqrt{3}(2\sqrt{3}) = 18+6 = 24 
eq 25$. For option $C$: $2(-1)+\sqrt{3}(9\sqrt{3}) = -2+27 = 25$. Thus,the normal passes through the point $(-1, 9\sqrt{3})$.

The normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$ at the point $(8, 3\sqrt{3})$ on it passes through the point

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